Solution 1 (Similar Triangles).

First observe that
$$\angle DPA = \angle BPC \stackrel{CP=CB}{=} \angle CBP = \angle CBD = \angle CAD = \angle PAD$$
so $DP = DA$. Thus there is a symmetry in the problem statement swapping $(A, D) \leftrightarrow (B, C)$.
Let $O$ be the centre of $\omega$ and let $E$ be the reflection of $P$ in $CD$ which, by
$$\angle CED = \angle DPC = 180^{\circ} - \angle CPB \stackrel{CP=CB}{=} 180^{\circ} - \angle PBC = 180^{\circ} - \angle DBC$$
lies on $\omega$. We claim the two lines concur at $E$. By the symmetry noted above, it suffices to prove that $BE \perp DQ$ and then $AE \perp CQ$ will follow by symmetry.
We have $AO = PQ$, $AD = DP$ and
$$\angle DAO = 90^{\circ} - \angle ABD \stackrel{PQ \perp AB}{=} \angle DPQ.$$
Hence $\triangle AOD \cong \triangle PQD$. Thus
$\angle QDB + \angle DBE = \angle ODA + \angle DAE \stackrel{DE=DA}{=} \angle ODA + \angle AED = (90^{\circ} - \angle AED) + \angle AED = 90^{\circ}$ giving $BE \perp DQ$ as required.

Solution 2 (Second Circle).

As in Solution 1, we prove that $DA = DP$ and note the symmetry in the problem statement swapping $(A, D) \leftrightarrow (B, C)$.
Let $\Gamma$ be the circumcircle of $\triangle PCD$. Since $DP = DA$ and $\angle ACD = \angle PCD$, the radius of $\Gamma$ is equal to that of $\omega$. We have that
$$\angle DPQ = \angle BPS = 90^{\circ} - \angle ABD = 90^{\circ} - \angle PCD.$$
This, combined with $PQ$ being equal to the common circumradius of $\Gamma$ and $\omega$, means that $Q$ is the circumcentre of $\Gamma$.
Let the perpendiculars to $CQ, DQ$ from $A, B$ intersect at $E$ then we have
$$\begin{aligned} & \angle EAC = 90^{\circ} - \angle ACQ \stackrel{QC=QP}{=} 90^{\circ} - \angle QPC = 90^{\circ} - \angle SPA = \angle CAB \Longrightarrow \angle EAB = 2\angle PAB \\ & \angle DBE = 90^{\circ} - \angle QDP \stackrel{QD=QP}{=} 90^{\circ} - \angle DPQ = 90^{\circ} - \angle BPS = \angle ABD \Longrightarrow \angle ABE = 2\angle ABP . \end{aligned}$$
Combining these
$$\angle BEA = 180^{\circ} - 2(\angle PAB + \angle ABP) = 180^{\circ} - 2\angle APD \stackrel{DA=DP}{=} \angle BDA$$
which gives that $E$ lies on $\omega$.