Much Synthetic Very Wow

We note that $D$ and $E$ are the midpoints of the tangent lines from $B$ to the incircle of $ABC$ . If we consider point $B$ to be a circle with radius $0$ , we get that $DE$ is the radical axis of $B$ and the incircle of $ABC$ . Thus, since $P$ lies on the radical axis of $B$ and the incircle, if $X$ is the tangency point of the incircle to $AC$ , then $PX = PB$ . Finally, by using Menelaus' Theorem, we compute $PA = \frac{21}{2}$ . Finally, we add $AX = 2$ to get $PX = PB = \frac{25}{2}$ , so the answer is $\boxed{666}.$