First, note that there is a lot of weak domination: $$Ll \succeq Lr, ~ Ll \succeq Rr, ~ Rl \succeq Rr, Lr \succeq Rr.$$$$Lr \succeq Ll, ~ Lr \succeq Rl, ~ Rr \succeq Rl, Ll \succeq Rl.$$
Consequently, $\nexists$ MSNE in which both players mix at the same time. If $P_1$ mixed, then $P_2$ would strictly prefer to play a pure strategy.
Consider then, as the answer suggests, a candidate equilibrium in which $P_1$ plays a pure strategy, whilst $P_2$ mixes.
If $P_1$ plays $T$, then $P_2$ is indifferent between $Ll$ and $Lr$.
$\implies$ $P_1 \to T, ~P_2 \to q(Ll) + (1 - q)Lr$.
$P_2$ is indifferent by construction, so we need merely show that $P_1$ prefers $T$ to $B$:
$$\begin{align} 3 & \geq 4q + 2(1 - q) \\
& \geq 2 + 2q \\
1 &\geq 2q \implies 1/2 \geq q
\end{align}$$
Likewise, if $P_1 \to T$, then it must be that $P_2 \to q(Lr) + (1-q)Rr$. Again, only need to show $P_1$ prefers $B$ to $T$: $$\begin{align} 2 & \geq 3q + 1(1 - q) \\ 1/2 & \geq q \end{align}$$
This fully describes the Nash Equilibrium.