Let $XYZ$ 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 $XY$, $C$ and $D$ on side $YZ$, and $E$ and $F$ on side $ZX$. Suppose $AB=CD=EF$. Find $ \angle XPY $ in degrees.
1. Given that $\angle X = 60^\circ$ and $\angle Y = 45^\circ$ in triangle $XYZ$, we can find $\angle Z$ using the fact that the sum of angles in a triangle is $180^\circ$:
   \[
   \angle Z = 180^\circ - \angle X - \angle Y = 180^\circ - 60^\circ - 45^\circ = 75^\circ
   \]

2. A circle with center $P$ passes through points $A$ and $B$ on side $XY$, $C$ and $D$ on side $YZ$, and $E$ and $F$ on side $ZX$. Given that $AB = CD = EF$, we can infer that $P$ is equidistant from the sides of the triangle $XYZ$.

3. Since $P$ is equidistant from the sides of the triangle, $P$ must be the incenter of triangle $XYZ$. The incenter is the point where the angle bisectors of the triangle intersect.

4. To find $\angle XPY$, we need to consider the properties of the incenter. The incenter divides each angle of the triangle into two equal parts. Therefore:
   \[
   \angle XPY = \frac{\angle X + \angle Y}{2}
   \]

5. Substituting the given angles:
   \[
   \angle XPY = \frac{60^\circ + 45^\circ}{2} = \frac{105^\circ}{2} = 52.5^\circ
   \]

The final answer is $\boxed{52.5^\circ}$