Let $K$ be the foot of the perpendicular from $D$ to $AB$ and $T$ the intersection of $AD$ with $k$. Since $\varangle A T B = \varangle A C B = 90^{\circ}$, $T$ also lies on the circumcircle of $ABC$. Because $C H \| D K$, it follows that $|A H| = |H K|$.
!

To prove that $C, Q$, and $T$ are collinear, we consider the intersection $Q'$ of $CT$ with $k$ and need only show that $P Q' k$ is tangent, or that $\varangle P Q' D = \varangle Q' B D$.
Since $B T Q' D$ is a cyclic quadrilateral and the triangles $A H C$ and $H K C$ are congruent, we have
$\varangle Q' B D = \varangle Q' T D = \varangle C T A = \varangle C B A = \varangle A C H = \varangle H C K$. Therefore, the right triangles $C H K$ and $B Q' D$ are similar, which implies $|H K| / |C K| = |Q' D| / |B D|$, hence $|H K| \cdot |B D| = |C K| \cdot |Q' D|$ (1). Since $P H \| D K$, we have $|P D| / |B D| = |H K| / |B K|$, thus $|P D| \cdot |B K| = |H K| \cdot |B D|$ (2). Comparing (1) and (2) leads to $|P D| \cdot |B K| = |C K| \cdot |Q' D|$ and thus $|P D| / |Q' D| = |C K| / |B K|$. Therefore, and because $\varangle C K A = \varangle K A C = \varangle B D Q'$, the triangles $C K B$ and $P D Q'$ are similar. Hence, $\varangle P Q' D = \varangle C B A = \varangle Q' B D$, which is what we needed to show.

Hint: A pure angle chase does not lead to the goal; however, there are other proof possibilities, such as utilizing a suitable homothety with center $B$.