Let $ Q $ be the intersection point of lines $ AD $ and $ BC $. From the condition it follows that $ \angle DQC=\angle AQB=60^{\circ}=\angle DPC $, so quadrilateral $ DPQC $ is cyclic, hence, $ \angle PDQ= \angle PCQ $, consequenty, $ \angle ADP=\angle BCP $, so triangles $ ADP $ and $ BCP $ are congruent, i.e. $ AP=BP $, and $ \angle APB={\angle APD}+{\angle BPD}={\angle BPC}+{\angle BPD}=60^{\circ} $. Thus, triangle $ APB $ is equilateral.

From this problem it's wise to remember the following old property: Take $M, N$ respectively on the sides $[AC], [AB]$ of the triangle ABC so that $BN=CM$. Then the perpendicular bisector of $[MN]$ passes through a fixed point. (midpoint of the arc $\stackrel{\frown}{BAC}$). The proof like above. Best regards, sunken rock