If you find the P1's utility of each pure strategy to P2 playing $(q_1,q_2,1-q_1-q_2)$, then you have:
$$ u_1 (H,(q_1,q_2,1-q_1-q_2)) = 2q_1 - 2q_2, \quad u_1(T, (q_1,q_2,1-q_1-q_2)) = -2q_1 + 2q_2, \quad u_1(O, (q_1,q_2,1-q_1-q_2)) = 3q_1 + 3q_2 -2.$$
Then setting all of these to be equal will give you $q_1 = q_2 = q_3 = \frac{1}{3}$.
However, this doesn't give you all the NE. Notice that when P2 plays $O$, P1 is indifferent between $H$ and $T$. So you might have a Nash equilibrium where P1 plays $p_1 H + (1-p_1) T$ and P2 plays $O$. We now to need to make sure that P2 playing $O$ is the best response to $p_1 H + (1-p_1) T$.
$$ u_2 ((p_1,1-p_1,0),H) = 2-4p_1, \quad u_1((p_1,1-p_1,0),T) = 4p_1 - 2, \quad u_2((p_1,1-p_1,0),O) = 1.$$
For $O$ to be the best response, we need $1 \ge 2-4p_1$ and $1 \ge 4p_1 - 2$, which together gives you $p_1 \in [\frac{1}{4},\frac{3}{4}]$.
So we have another set of equilibrium which look like $$\left\{ (p_1 H + (1-p_1)T,O), \frac{1}{4} \le p_1 \le \frac{3}{4} \right\}.$$
Apply the same logic to the other player to get $$\left\{ (O, p_2 H + (1-p_2)T), \frac{1}{4} \le p_2 \le \frac{3}{4} \right\}.$$