BAMO

Solution:

It will suffice to show that $\angle AQR = \angle CQP$ (compare with Problem 3.) Note that both quadrilaterals $MQAR$ and $MQPC$ are cyclic: $\angle MRA = 90^{\circ} = \angle MQA$, and $\angle MQC = 90^{\circ} = \angle MPC$. Using inscribed angles in these quadrilaterals, we obtain that the two angles in question equal correspondingly to:
$$\angle AQR = \angle AMR, \quad \angle CQP = \angle CMR.$$
Look carefully at $\triangle MAR$ and $\triangle MCP$: both have one right angle, and we hope to show that two more angles are the same. Equivalently, we want to show that these two triangles are similar. Can we do that? Yes. Their third angles are the same: $\angle MCP = \angle MAR$ because $\angle MAR = 180^{\circ} - \angle MAB = \angle MCB$ from the inscribed quadrilateral $MCBA$.

Thus, we conclude that $\triangle MAR \sim \triangle MCP$ (two same angles), so their third angles are also the same: $\angle CMP = \angle RMA$. Combining this with (1), we finally obtain $\angle AQR = \angle CQP$, and therefore $P, Q, R$ are collinear.