SOLUTION:
1) DEFINITION OF EQUILIBRIUM
\begin{equation} \begin{split}
&p \in \mathbb{R}_+\text{ is the price chosen by the firm in period 1}\\
&a_U \in [0,1] \text{ is the share of income the Uninformed consumer allocates in period 1}\\
&a_I \in [0,1] \text{ is the share of income the Informed consumer allocates in period 1}\\
&\Theta = \{I,U\} \text{ is the type of consumer: Informed, Uninformed}\\
& S_I = \{ a : \{\underline{p},\bar{p}\} \times \mathbb{R}_+ \to [0,1] \} \text{ is the strategy set of the I. consumer}\\
& S_U = \{ a : \mathbb{R}_+ \to [0,1] \} \text{ is the strategy set of the U. consumer}\\
& S_f = \{ p : \{\bar{p},\underline{p}\} \to \mathbb{R}_+ \} \text{ is the strategy set of the firm}\\
&\mu_U(p) = Pr(P =\bar{p}|p) \text{ : consumer of type U belief that the second period price is } \bar{p}, \text{ given } p\\
&\mu_I(P) = Pr(P =\bar{p}|p) \text{ : consumer of type I belief that the second period price is } \bar{p}, \text{ given } p\\
\end{split} \end{equation}
An equilibrium of the game is a triple of strategies, $s^*=(a^*_I, a^*_U,p^*)$ for each type of consumer and the monopolist, together with a system of belief $\mu$ such that: for each player $i$, $s_i^*$ is sequentially rational, and the beliefs are determined by the strategy profile and Bayes rule whenever it applies. \
In particular for each price $p$, the consumer of type $i \in \{U.I\}$ solves:
\begin{equation} \begin{split}
&a_i^*(p) = \text{ argmax } \sum_p \big[ ( \frac{a}{p} + \frac{1-a}{\bar{p}} )\mu_I(P) + ( \frac{a}{p} + \frac{1-a}{\underline{p}} )(1-\mu_I(P)) \big]
\end{split} \end{equation}
Also, in the case of the Informed consumer, we know that
\begin{equation} \begin{split}
&\mu_I(P) = Pr(P =\bar{p}|p) =
\begin{cases}
1 &\text{ if } P =\bar{p}\\
0 &\text{ if } P = \underline{p}\\
\end{cases} \\
\end{split} \end{equation}
given the consumers optimal strategies and $P$, $p^*(P)$ maximizes the monopolist first period profits:
\begin{equation} \begin{split}
&\pi(P)^* = \text{ argmax } \alpha [a^*_I(p) p] + (1-\alpha) [a^*_U(p) p]\\
\end{split} \end{equation}
2) CHARACTERIZATION OF EQUILIBRIA
for any realization of $P$ in period 2, we have that
\begin{equation} \begin{split}
a_I(p)^* =
\begin{cases}
1 &\text{ if } p < P\\
0 &\text{ if } p >P\\
\in [0,1] &\text{ if } p = P \\
\end{cases} \\
\end{split} \end{equation}
\begin{equation} \begin{split}
a_U(p)^* =
\begin{cases}
1 &\text{ if } p < \mu_U(p) \bar{p} + (1-\mu_U(p))\underline{p}\\
0 &\text{ if } p > \mu_U(p) \bar{p} + (1-\mu_U(p))\underline{p}\\
\in [0,1] &\text{ if } p = \mu_U(p) \bar{p} + (1-\mu_U(p))\underline{p}
\end{cases} \\
\end{split} \end{equation}
To define the optimal strategy of the monopolist consider the following cases, where $\pi$ denotes the monopolist first period profits:
a)Suppose $p=\bar{p}, \forall P$, then By Bayes rule $\mu_U(p) = 1/2$; also suppose that off the equilibrium path $\mu_U(p) = 1/2, \forall p \neq\bar{p}$ ; then the monopolist profits are given by:
\begin{equation} \begin{split}
\pi(P) =
\begin{cases}
\alpha \bar{p} a_I^*(\bar{p}) &\text{ if } P=\bar{p} \\
0 &\text{ if } P = \underline{p} \\
\end{cases} \\
\end{split} \end{equation}
This is not an equilibrium strategy because the monopolist can deviate and increase its profits when $P=\underline{p}$ by setting, for example, $\hat{p} = 1/2\bar{p} + 1/2\underline{p}$; this will give him a profit of $\alpha a_U^*(\hat{p}) >0$.
Similarly it is possible to prove that no-matter what the off the equilibrium path are, there is always a profitable deviation for the monopolist. \
b)Suppose $p=\underline{p}, \forall P$, then $\mu_U(p) = 1/2$ and also suppose that off the equilibrium path $\mu(p) = 1/2, \forall p \neq\underline{p}$ ; then the monopolist profits are given by:
\begin{equation} \begin{split}
\pi(P) =
\begin{cases}
\underline{p} &\text{ if } P=\bar{p} \\
\underline{p} (\alpha a_I^*(\underline{p}) +(1-\alpha) ) &\text{ if } P = \underline{p}\\
\end{cases} \\
\end{split} \end{equation}
this is not an equilibrium strategy because $\exists \epsilon > 0 $ such that monopolist will deviate and increase its profits when $P=\bar{p}$; for example by setting $\hat{p} = 1/2\bar{p} + 1/2\underline{p} -\epsilon>\underline{p}$ which will give him a profit of $\hat{p}> \underline{p}$ .
c)Suppose $p=\underline{p}, \text{ if } P= \underline{p}$, $p=\bar{p}, \text{ if } P= \bar{p}$ and $\mu_U(p) = 1, \forall p > \underline{p}$ then
\begin{equation} \begin{split}
\pi_1^*(P) =
\begin{cases}
\bar{p}(\alpha a_I^*(\bar{p}) +(1-\alpha) a_U^*(\bar{p})) &\text{ if } P=\bar{p} \\
\underline{p} (\alpha a_I^*(\underline{p}) +(1-\alpha) a_U^*(\underline{p})) &\text{ if } P = \underline{p}\\
\end{cases} \\
\end{split} \end{equation}
this represents an equilibrium as long as $a_I^*(\bar{p}) = a_U^*(\bar{p}) = 1$; otherwise the monopolist can set a price $\hat{p}< \bar{p}$ and can achieve a profit of $\bar{p} - \epsilon >\alpha a_I^*(\bar{p}) +(1-\alpha) a_U^*(\bar{p})$
d) Suppose $p = 1/2\bar{p} + 1/2\underline{p}, \forall P$ then $\mu_U(p)=1/2$ and and also suppose that off the equilibrium path $\mu(p) = 1/2, \forall p \neq ( 1/2\bar{p} + 1/2\underline{p}) $
\begin{equation} \begin{split}
\pi_2^*(P) =
\begin{cases}
(1/2\bar{p} + 1/2\underline{p}) (\alpha +(1-\alpha) a_U^*(p)) &\text{ if } P=\bar{p} \\
( 1/2\bar{p} + 1/2\underline{p} ) ( 0 + (1-\alpha)a_U^*(p)) &\text{ if } P = \underline{p}\\
\end{cases} \\
\end{split} \end{equation}
This is an equilibrium strategy as long as there are no profitable deviation; notice that the monopolist can set $p=\bar{p}$ when $P=\bar{p}$ and given $\mu_U = 1/2$ its profits would equal $\alpha \bar{P}a_I^*(\bar{p})$. Also, when $P=\underline{p}$ the monopolist could set $p=\underline{p}$ and get a profit of $\underline{p}$. So we need to insure that neither deviation is possible.
e) ?
We can therefore characterize the following (Perfect Bayesian) Equlibria:
1ST EQUILIBRIUM)
\begin{equation} \begin{split}
&a_I(p)^* = 1, a_U(p)^* = 1 \\
&\pi_1^*(P) \text{ defined as above } \\
&\mu_U(p) = 1 \text{ both "on and off" the equilibrium path } \\
&\text{(that is, if the monopolist deviates } \\
&\text{from the equilibrium strategy } \mu_U(p)=1 \text{ still holds} \\
\end{split} \end{equation}
SET OF EQUILIBRIA)
\begin{equation} \begin{split}
&a_I(p)^* =1, a_U(p)^* \in[0,1] \\
&\pi_2^*(P) \text{ defined as above } \\
&\mu_U(p) = 1/2 \text{ both "on and off" the equilibrium path } \\
&\text{(that is, if the monopolist deviates } \\
&\text{from the equilibrium strategy } \mu_U(p)=1/2 \text{ still holds} \\
& \text{Also, The following two conditions below hold:}\\
&\alpha \bar{P}a_I^*(\bar{p}) < (1/2\bar{p} + 1/2\underline{p}) (\alpha a_U^*(\bar{p}) +(1-\alpha) ) \\
&\underline{p} < ( 1/2\bar{p} + 1/2\underline{p} ) (\alpha a_U^*(\bar{p}) +0)\\
\end{split} \end{equation}