Solution:

Note that $\triangle QAB \sim \triangle QDC$ (by corresponding angles). Since $QM$ and $QN$ are corresponding medians, we have $\angle AQ M = \angle DQ N$ and hence $Q$, $M$, and $N$ are collinear.

Similarly, note that $\triangle PAB \sim \triangle PCD$. Since $PM$ and $PN$ are corresponding medians, we have $\angle AP M = \angle CP N$ and hence $P$, $M$, and $N$ are collinear. Since $P$ and $Q$ lie on line $MN$, all four of these points are collinear.