This sounds like homework. Here's a hint: use the inscribed angle theorem, which states two things of interest to you:
- The angle $\angle ABC$ inscribed within a circle by points $A$, $B$, and $C$ located on the edge of the circle is always exactly half of the central angle of the intercepted arc between B and C.
- Opposing vertices of any quadrangle that can be inscribed by a circumcircle (a cyclic quadrangle) have supplementary angles.
These allow you to form the following lemmas:
- The arcs $AM$ and $MC$ each have the same central angle measure as $\angle ABC$.
- The angles $\angle EDC$ and $\angle EBC$ are supplementary, and the angles $\angle BED$ and $\angle BCD$ are supplementary, because by definition they form a quadrangle within the circumcircle of $\triangle BCD$.
- By the same token, angles $\angle PBC$ and $\angle PMC$ are supplementary, and $\angle BPM$ and $\angle BCM$ are supplementary.
Here's something that looks right to me, but my geometry is insufficient to prove it; I think if you can, it'll go a long way:
- Quadrangles $BCDE$ and $PMCB$ are similar; $BCDE$ is upside-down relative to $PMCB$. Prove this, and you get $\angle ABC \cong \angle PMC$ and $\triangle PMC \cong \triangle ABC$, therefore $MC = BC$.