Saudi Arabian Mathematical Competitions

Because quadrilateral $BPDQ$ is cyclic, we have $\angle PQD = \angle PBD$. Because quadrilateral $ACQP$ is cyclic, we have $\angle PQC = 180^{\circ} - \angle CAP$. We deduce that
$$\begin{aligned} \angle DQC & = \angle PQC - \angle PQD = 180^{\circ} - \angle CAP - \angle PBD \\ & = 180^{\circ} - \angle XAB - \angle ABX = \angle BXA = \angle DXA. \end{aligned}$$
This proves that quadrilateral $CQDX$ is cyclic and therefore
$$\angle XQC = \angle XDC.$$
Because $PA$ is parallel to $DX$, we have $\angle YPC = \angle XDC$. Because $CQPY$ is cyclic, we have $\angle YPC = \angle YQC$. Therefore, $\angle YQC = \angle XQC$, which means that points $Q, X, Y$ are collinear.

Because quadrilateral $BPQZ$ is cyclic, we have $\angle ZQB = \angle ZPB$. Because lines $PZ$ and $AC$ are parallel, we have $\angle ZPB = \angle CAP$.
Let $C'$ be the second intersection point of line $CQ$ with circle $\omega_{1}$. Because quadrilateral $ACQP$ is cyclic, we have $\angle C'QP = \angle CAP$. Therefore, $\angle ZQB = \angle C'QP$. We deduce, by cyclicity of quadrilaterals $BPDQ$ and $CQDX$, that
$$\angle ZQC' = \angle BQP = \angle BDP = \angle XDC = \angle XQC,$$
which means that points $Q, X, Z$ are collinear.