WLOG, assume $AD<BC$. Since, $\Delta ADO \sim \Delta CBO$ together with the parallel condition of $MP,MQ$ implies, $$\frac{AO}{CO}=\frac{AD}{BC}=\frac{CM \cdot DQ}{MD \cdot CP}$$Recall, $\frac{DQ}{AQ}=\frac{MD}{CM}$, hence, $$\frac{AO}{CO}=\frac{AD}{BC}=\frac{CM \cdot DQ}{MD \cdot CP}=\frac{AQ}{CP} \implies \Delta AQO \sim \Delta CPO \implies P-O-Q$$

Just to show some details that may be a bit unclear in @aboves solution, we get that $\frac{AO}{CO}=\frac{AD}{BC}=\frac{\frac{AD}{CD}}{\frac{BC}{CD}} =\frac{\frac{DQ}{MD}}{\frac{CP}{CM}}=\frac{CM\cdot DQ}{MD\cdot CP}$. Also, does anybody have solution using Desargue? It is under Desargue section of ABJTOG, so was just wondering...

@above Desargue is most often hidden intrinsically under other projective theorems. In this case it's just Pappus with line at infinity.

The points $P, Q$ are the corresponding points of similar triangles $\triangle BOC, \triangle DOA$. So $\angle COP=\angle AOQ$, so $P, O, Q$ are collinear.