Solution:

First solution. Since $CP = CQ$, $CA = CB$ and $\angle APC = \angle BQC = 90^\circ$, then $\triangle APC \cong BQC$. Hence $\angle CAP = \angle CBQ$. Setting $AP \cap BQ = T$, it follows that the quadrilateral $ABTC$ is cyclic. Then $\angle BAC = \angle QTC$ and now $\angle TQC = \angle AHC = 90^\circ$ implies that $\angle QCT = \angle ACH$. The equalities $\angle AHC = \angle APC = \angle CPT = \angle CQT = 90^\circ$ show that $AHPC$ and $CPTQ$ are cyclic quadrilaterals. Thus $\angle APH = \angle ACH$ and $\angle QPT = \angle QCT$ which means that $\angle APH = \angle QPT$. Hence the points $H$, $P$ and $Q$ are collinear.

Second solution. Set $S = HQ \cap k$. Since the quadrilateral $BHCQ$ is cyclic, and the triangles $ABC$ and $CQS$ are isosceles, it follows that $\angle BAC = \angle ABC = \angle HQC = \angle CSQ$. Then $AHSC$ is a cyclic quadrilateral and therefore $\angle ASC = \angle AHC = 90^\circ$. Hence $S = P$, i.e., the points $H$, $P$ and $Q$ are collinear.