Mixed Strategies in Bayes Nash Equilibrium (Bayesian Battle of the Sexes). Shouldn't it depend on $p$?
Suppose player 1 chooses a mixed strategy such that the left type of player 2 is indifferent between C and H. There is a unique mixed strategy that does this. The same mixed strategy, however, does NOT make the right type of player 2 indifferent. Loosely speaking, this is simply because the preferences of the left type are exactly those of the right type switched around. Thus it is impossible for player 1 to make both types indifferent simultaneously. Therefore, the probability $p$ plays no role when choosing a strategy such that at least one of them is indifferent.
This also indicates why there are exactly two equilibria.