Solution:

The answer is (C). Let $M$ be the midpoint of $AB$. The angles $\widehat{PBM}$ and $\widehat{PCM}$ are congruent because they subtend the same arc. Since the triangle is isosceles, the angles $\widehat{ABC}$ and $\widehat{ACB}$ are equal and therefore, by subtracting equal angles, $\widehat{MCB}=\widehat{PBC}$ and $M$ is also the foot of the bisector of the angle $\widehat{ACB}$, just as $P$ is also the midpoint of the side $AC$, since the two triangles $MBC$ and $BCP$ are congruent. Since $CM$ is both a median and a bisector, the triangle $ABC$ is isosceles at $C$, that is, the sides $BC$ and $AC$ are equal. Then $ABC$ is equilateral and the angle sought has measure equal to $60^\circ$.