The angle $\varangle P D C$ is on the one hand the adjacent angle of $\varangle A D P$, and on the other hand, it lies in the cyclic quadrilateral $B P D C$ opposite the angle $\varangle C B P$. Consequently, $\varangle A D P = \varangle C B P$. Similarly, the angle $\varangle C M P$ is the adjacent angle of $\varangle P M B$ and lies in the cyclic quadrilateral $A P M C$ opposite the angle $\varangle P A C$, so $\varangle P M B = \varangle P A C$. The triangles $P B M$ and $P D A$ thus agree in the interior angles at $B$ and $D$ as well as at $M$ and $A$, and furthermore, by assumption, $\overline{A D} = \overline{A M} = \overline{M B}$ (the last equality follows from Thales' theorem). According to the congruence theorem wsw, the triangles $P B M$ and $P D A$ are congruent, in particular, their heights from $P$ are of equal length. They are, however, the perpendiculars from $P$ to the triangle sides $C A$ and $C B$, so $P$ lies on the angle bisector.