Solution:

The angle $\text{P D C}$ is on the one hand the supplementary angle of $\text{A D P}$, and on the other hand, in the cyclic quadrilateral $B P D C$, it is opposite the angle $\text{C B P}$. Hence $\text{A D P = C B P}$. Analogously, the angle $\text{C M P}$ is the supplementary angle of $\text{P M B}$ and, in the cyclic quadrilateral $A P M C$, it is opposite the angle $\text{P A C}$, so $\text{P M B = P A C}$. 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 $A$; moreover, by assumption $\overline{A D} = \overline{A M} = \overline{M B}$ (the last equality follows from Thales' theorem). By the ASA congruence theorem, the triangles $P B M$ and $P D A$ are congruent; in particular, their altitudes from $P$ have equal length. But these are the perpendiculars from $P$ to the triangle sides $C A$ and $C B$, respectively, so $P$ lies on the angle bisector.

Solution:

Denote by $m_{X Y}$ the perpendicular bisector of the segment $X Y$. Then both $m_{C D}$ and $m_{C A}$ as well as $m_{C M}$ and $m_{C B}$ are pairs of parallel lines at distance $\frac{1}{4} \overline{B C}$, so they bound a rhombus. The centers $M_{1}, M_{2}$ of the two circumcircles mentioned in the problem are the intersection points of $m_{C A}$ with $m_{C M}$ and of $m_{C D}$ with $m_{C B}$, respectively. The line $(C P)$ is perpendicular to the connecting line $\left(M_{1} M_{2}\right)$, hence it is parallel to the second diagonal of the rhombus. One easily convinces oneself that the second diagonal is parallel to the angle bisector.