First observe that
$$\angle DPA = \angle BPC = \angle CBP = \angle CBD = \angle CAD = \angle PAD$$
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
$$\angle CED = \angle DPC = 180^\circ - \angle CPB = 180^\circ - \angle PBC = 180^\circ - \angle DBC$$
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
$$\angle DAO = 90^\circ - \angle ABD = \angle DPQ.$$
Hence $\triangle AOD \cong \triangle PQD$. Thus
$$\angle QDB + \angle DBE = \angle ODA + \angle DAE = \angle ODA + \angle AED = 90^\circ$$
giving $BE \perp DQ$ as required.

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
$$\angle EAC = 90^\circ - \angle ACQ = 90^\circ - \angle QPC = 90^\circ - \angle SPA = \angle CAB \Rightarrow \angle EAB = 2\angle PAB$$
$$\angle DBE = 90^\circ - \angle QDP = 90^\circ - \angle DPQ = 90^\circ - \angle BPS = \angle ABD \Rightarrow \angle ABE = 2\angle ABP.$$
Combining these
$$\angle BEA = 180^\circ - 2(\angle PAB + \angle ABP) = 180^\circ - 2\angle APD = \angle BDA$$
which gives that $E$ lies on $\omega$.