Solution:

a. The quadrilateral $ANZM$ is a parallelogram because the diagonals intersect at their midpoint by construction. The lines $AN$ and $MZ$ are therefore parallel. On this pair of parallel lines, the transversals $CN$ and $CA$ cut off equal corresponding angles ($\angle CMP = \angle CNA$ and $\angle CPM = \angle CAN$). It remains only to observe that $CA = CN$ because they are radii of the circle $\gamma$: from this it follows that the triangle $NCA$ is isosceles on $NA$, hence that the base angles $\angle CAN = \angle CNA$ are equal. In conclusion, we know that
$$\angle CPM = \angle CAN = \angle CNA = \angle CMP$$
that is, that $\triangle MCP$ is isosceles on $MP$.

b. The triangle $BMX$ is isosceles because $BM = BX$ are radii of the circle $\omega$, hence the base angles $\angle BXM = \angle BMX$ are equal. Moreover, $\angle BMX$ is vertically opposite to $\angle CMP$, and they are therefore equal to each other. It follows, by point (a), that all the angles in question are also equal to $\angle MPC$:
$$\angle BXM = \angle BMX = \angle CMP = \angle MPC.$$
The equality between the two extremes tells us that the transversal $PX$ cuts off on the pair of lines $BX$ and $AC$ equal alternate interior angles, thereby proving their parallelism.

c. The triangle $ABX$ is isosceles because $BA = BX$ are radii of the circle $\omega$, hence the base angles $\angle BAX = \angle BXA$ are equal. The transversal $AX$ cuts off on the pair of lines $BX \parallel AC$, parallel by point (b), equal alternate interior angles $\angle BXA = \angle CAX$. We have obtained that
$$\angle BAX = \angle BXA = \angle CAX$$
that is, that $X$ lies on the bisector of the angle $\angle BAC$. Repeating the same reasoning for the point $Y$, with the appropriate changes of letters, we obtain that it too lies on the bisector of the angle $\angle BAC$, which gives the thesis.