Figure 2

Choose the point $K$ such that $A B K C$ is a square. Let $N$ be the point of intersection of $A P$ and $B K$ (see Figure 2). Since the lines $A N$ and $C M$ are perpendicular, $N$ is the midpoint of $B K$. Moreover, triangles $A M C$ and $B N A$ are congruent, which gives
$$\angle A M C=\angle B N A$$
Since $|B M|=|B N|$ and $\angle M B P=\angle N B P$, it follows that triangles $M B P$ and $N B P$ are congruent. This implies that
$$\angle B M P=\angle B N P$$
Combining (1) and (2) yields the required equality.