Extend $AP$ to meet $BC$ at point $D$; then $P$ is the midpoint of $AD$. Since $M$ is the midpoint of $AC$, $PM$ is parallel to $CD$ (i.e., $BC$), and
$$\angle QPM = \angle QBC = \frac{1}{2} \angle ABC.$$
By the same reasoning, $\angle MQP = \frac{1}{2}\angle ABC$, so $PM = QM$.

$$\angle QHC = \angle QHM = \angle QPM,$$
that is, $\angle QHC = \angle QBC$. Hence $Q,H,B,C$ are concyclic, and
$$\angle BHC = \angle BQC = 90^\circ,$$
so we get $HN = \frac{1}{2}BC = NQ$.
Since $OH = OQ$, $ON$ is the perpendicular bisector of segment $HQ$. Also, since $\angle MQP = \frac{1}{2}\angle ABC$ and $N$ is the midpoint of $BC$, we obtain that $Q,M,N$ are three collinear points. Therefore
$$\angle NHO = \angle NQO = \angle MQO = \angle OMQ,$$
which shows that $O,M,N,H$ are concyclic. This completes the proof.