Since $OP \parallel AB$ and $P$, $C$, $O$, $Q$, $B$ are cyclic, we have $\angle POC = \angle PQC = \angle PBC = \alpha$ and $\angle POB = \angle PCB = \angle PQB$. It follows that $\triangle OAB \sim \triangle BPC$. Hence
$$\frac{BP}{PC} = \frac{OA}{OB}.$$
Because $ABCD$ is a trapezoid, we have
$$\frac{OC}{OA} = \frac{OD}{OB} \iff 1 + \frac{OC}{OA} = \frac{OC}{OB} + 1 \iff \frac{AC}{BD} = \frac{OA}{OB}$$
From this, we deduce that
$$\frac{BP}{PC} = \frac{AC}{BD}.$$
On the other hand, we have $\frac{QC}{QB} = \frac{AC}{DB}$ since $\triangle AQC \sim \triangle DQB$. Consequently,
$$\frac{QC}{QB} = \frac{BP}{PC}. \text{ Now by the law of sines}$$
$$\frac{CM}{MB} = \frac{CM}{MP} \cdot \frac{MP}{MB} = \frac{\sin \angle CPM}{\sin \angle MCP} \cdot \frac{\sin \angle CBP}{\sin \angle MPB} = \frac{CQ}{PB} \cdot \frac{CP}{BQ} = 1.$$

Thus $CM = MB$.