Let $\measuredangle APC=2\alpha$ and $\measuredangle BQD=2\beta$; then we have $$180^{\circ}-\measuredangle DAB=\left(\measuredangle IPQ-\alpha\right)+\left(\measuredangle IQP-\beta \right),$$hence, $$\measuredangle DAB=\measuredangle PIQ-(\alpha+\beta).$$Analogously, we may conclude that $$\measuredangle DCB=\measuredangle PIQ+(\alpha+\beta).$$Consequently, we have $$2\measuredangle PIQ=\measuredangle DAB+\measuredangle DCB=180^{\circ} \implies \measuredangle PIQ=90^{\circ}.$$Finally, by Brokard's Theorem, we know that $\odot(PQ)$ and $\odot(O)$ are orthogonal, so we are done.

A shorter way to show $\angle PIQ =90^\circ$: Notice that $PI$ corresponds to the bisector in both $PAD$ and $PBC$, so it makes equal angles with $AB, CD$, so $QI$ is the angle bisector in an isosceles triangle and hence we conclude. Note: My solution in the contest used Newton's theorem (touchchords meet at intersection of diagonals for a quad with an incircle) and the fact that $PI$ is perp to a touchchord and that the diagonals of the in touch quadrilateral of a bicentric quadrilateral are perpendicular, so $\angle PIQ = 90^\circ$ by noticing a rectangle.

Let $O$ be the center of $\Gamma$ and $R=AC\cap BD.$ It's well known $\angle PIQ=90^\circ$ and $O,P,Q,R$ form an orthocentric quadruple. Now let $X=OR\cap PQ,Y=OP\cap QR$ and observe that $$\text{pow} (O,\odot (PIQ))=|OY|\cdot |OP|=|OR|\cdot |OX|=|OA|^2.$$