 Solution

Write $p(s)=p(T,s;K_3,T_3)$ and $c(s)=c(T,s;K_3,T_3)$. Because the right can be left unexercised, the chooser with positive exercise cost has payoffLet $\tau =T_3-T$ and $D=K_3{e}^{-r\tau }$. Put-call parity gives $c(s)-p(s)=s-D$. Thus the put is selected for $s<D$ and the call for $s>D$; the two are equal at $s=D$. Before truncation at zero, the payoff is decreasing on the left and increasing on the right.

At the meeting point, the Black-Scholes formula givesIf $K<M$, the payoff has a positive minimum $M-K$ at $D$. If $K=M$, it touches zero there. If $M<K<D$, each side has a distinct exercise threshold and the payoff is zero between them. If $K\ge D$, even the left-end put value cannot cover the exercise cost, and only the right call branch can be positive. The illustration shows the middle case, including the competing underlying option values.