Let $E$ be the midpoint of the arc $BA$ that doesn't contain $C$, let $F$ be the midpoint of the arc $DA$ that doesn't contain $C$ and let $G=PE\cap QF$. Clearly $A$ and $P$ are symmetric with respect to $CE$, and $A$ and $Q$ are symmetric with respect to $CF$, so\begin{align*}\angle CEG & =180^\circ -\angle PEC \\ & =180^\circ -\angle CEA \\ & =\angle AFC \\ & =\angle CFQ, \end{align*}and therefore $GECF$ is cyclic. We are done by Pascal on $EGFBCD$.

Let $ABCD$ be inscribed in the unit circle, and let $X, Y$ be the midpoints of arcs $AB$ and $AD$ (not containing $C$) respectively. Then we have $$|a|=|c|=|x|=|y|=1$$$$b = \frac{x^2}a$$$$d = \frac{y^2}a$$Since $P$ is the reflection of $A$ over line $CX$, and $Q$ is the reflection of $A$ over line $CY$, we have $$p = c + x - \frac{cx}a$$$$q = c + y - \frac{cy}a$$Let $J$ be the incenter of $\triangle ABD$. Then $$j = \frac{by(d+x) - dx(b+y)}{by-dx} = x + y - \frac{xy}a$$and we have $$\frac{p-j}{q-j} = \frac{c-y-\frac{cx}a+\frac{xy}a}{c-x-\frac{cy}a+\frac{xy}a} = \frac{(a-x)(c-y)}{(a-y)(c-x)}$$which is real. So $J$ lies on line $PQ$. $\blacksquare$

Let $I$ be the incenter of $ABD$. By angle chasing, it is easy to show that $(A,I,D,Q)$ and $(A,I,B,P)$ are concyclic. Therefore, $\angle AIP=\angle ABP=180^{\circ}-\angle ADQ= 180^{\circ}-\angle AIQ$, so $(P,I,Q)$ collinear. $\blacksquare$

Let circles $(AQD)$ and $(APB)$ intersect again at $I$. Claim: $P$, $I$, and $Q$ are collinear $$\angle AIQ+\angle AIP=\angle ADQ+\angle ABP=180^{\circ}$$Claim: $I$ is the incenter of $ABD$ $$\angle AID=180^{\circ}-\angle DQA=90^{\circ}+\frac{1}{2}\angle ACD=90^{\circ}+\frac{1}{2}\angle ABD$$$$\angle AIB=180^{\circ}-\angle BPA=90^{\circ}+\frac{1}{2}\angle ACB=90^{\circ}+\frac{1}{2}\angle ADB$$

it is an easy question)

Let $I' = (AQD) \cap (APB)$. Then one can find angles $AI'D$ and $AI'B$ using angle chasing, yielding that $I'$ is the incentre of $ABD$, from where some more angle chasing yields that $\angle AI'Q + \angle AI'P = 180^\circ$, which finishes. $\square$