Solution:
$$
255 / 2
$$

Since $P A B, P C D$, and $P E F$ are all isosceles triangles with equal legs and equal bases, they are congruent. It follows that the heights of each are the same, so that $P$ is equidistant from the sides of $X Y Z$. Therefore, $P$ is the incenter and therefore lies on the angle bisectors of $X Y Z$. Thus $\angle Y X P=\frac{1}{2} \angle Y X Z=30^{\circ}$ and $\angle P Y X=\frac{1}{2} \angle Z Y X=$ $\frac{45^{\circ}}{2}$. Therefore $\angle X P Y=180^{\circ}-30^{\circ}-\frac{45^{\circ}}{2}=\frac{255^{\circ}}{2}$.
4. Let $X Y Z$ be a triangle with $\angle X=60^{\circ}$ and $\angle Y=45^{\circ}$. A circle with center $P$ passes through points $A$ and $B$ on side $X Y, C$ and $D$ on side $Y Z$, and $E$ and $F$ on side $Z X$. Suppose $A B=C D=E F$. Find $\angle X P Y$ in degrees.