Pure-Strategy Bayesian Nash equilibrium with general common prior

The best way to visualise what's going on would be to use Harsanyi's transformation. I'm not drawing the game tree here (but I think Tirole has it in his example).

Let's set up notations first. We will denote player 1's strategy by $x=Pr(T)$. We will call the decision of player 2 following the realisation of game $i$ by $y_i=Pr(T)$ - i.e. player 2, following realisation of game A, chooses T with probability $y_A$.

Simple calculation of the best responses give us the following expressions: $y_A=begin{cases} 0 & text{ if } x<frac{1}{4}\ [0,1] & text{ if } x=frac{1}{4} \ 1 & text{ if } x>frac{1}{4}\ end{cases}$

$y_B=begin{cases} 0 & text{ if } x>frac{1}{4}\ [0,1] & text{ if } x=frac{1}{4} \ 1 & text{ if } x<frac{1}{4}\ end{cases}$

$x=begin{cases} 0 & text{ if } rho[4y_A-3] +(1-rho)[4y_B-3]< 0\ [0,1] & text{ if } rho[4y_A-3] +(1-rho)[4y_B-3]=0\ 1 & text{ if } rho[4y_A-3] +(1-rho)[4y_B-3]> 0\ end{cases}$

The set of BNE of the above game is the tuple $(x,y_A,y_B)in[0,1]^3$ that satisfies the above three equations. The solution is quite simple:

1. For any $rhoin[0,1]$, $x=frac{1}{4},;y_A=y_B=frac{3}{4}$ is an equilibrium.

2. For $rhoin[frac{1}{4},1]$, $x=0,;y_A=0,;y_B=1$ is an additonal equilibrium to the above.

3. For $rhoin[frac{3}{4},1]$, $x=1,;y_A=1,;y_B=0$ is an additonal equilibrium to the above two.

I hope there are no calculation errors, but the key idea should remain the same.

Matrix looks correct. To final all pure strategy BNEs, you'll have to discuss cases based on the value of $p$.

For example, if $pin(0,1)$, then $FT$ is player 2's unique best response to $F$. Thus, to have a BNE, you'd want $F$ to be player 1's best response to $FT$ as well, meaning that you'd require $3p>1-p$, or $p>frac14$. Hence, $(F,FT)$ is a BNE if $pin(frac14,1)$.

You should be able to find other BNEs following a similar line of reasoning. (Hint: don't forget the edge cases where $p=0$ and $p=1$.)