The formulation was kind of dumb imho. We just want to prove $PQ$ is radical axis of $(AQC)$ and $(BQD)$. Let $(AQC)$ meet $PD$ at $D'$ and let $(BQD)$ meet $PA$ at $A'$. Proving $A'D'DA$ is cyclic is sufficient by PoP. It is equivalent to $A'D'CB$ being cyclic by Reim's. Now, this is an easy angle chase: $\angle BA'C = \angle CQD= \angle AQB= \angle BD'C$, so we are done.

Bashed in ~15 min during the contest: We employ complex numbers. Let $\overline{AD}$ be the real axis and $Q=0$. Let $M$ be the midpoint of $BC$. Then $BC\perp MO\perp AD$ because $QB=QC$, so $c=m+x$ and $b=m-x$ where $x\in\mathbb{R}$ and $m\in i\mathbb{R}$. Let $O_{1}$ be the circumcenter of $\triangle AQC$ and $O_{2}$ be the circumcenter of $\triangle DQB$. Now we compute $o_{1},o_{2},p$: \[o_{1}=\frac{\begin{vmatrix} a & a\overline{a} & 1\\ q & q\overline{q} & 1\\ c & c\overline{c} & 1 \end{vmatrix}}{\begin{vmatrix} a & \overline{a} & 1\\ q & \overline{q} & 1\\ c & \overline{c} & 1 \end{vmatrix}}=\frac{\begin{vmatrix} a & a^2 & 1\\ 0 & 0 & 1\\ m+x & (m+x)(-m+x) & 1 \end{vmatrix}}{\begin{vmatrix} a & a & 1\\ 0 & 0 & 1\\ m+x & -m+x & 1 \end{vmatrix}}=\frac{a^2(m+x)-a(m+x)(-m+x)}{a(m+x)-a(-m+x)}=\frac{(m+x)(a+m-x)}{2m}\]\[o_{2}=\frac{\begin{vmatrix} d & d\overline{d} & 1\\ q & q\overline{q} & 1\\ b & c\overline{b} & 1 \end{vmatrix}}{\begin{vmatrix} d & \overline{d} & 1\\ q & \overline{q} & 1\\ b & \overline{b} & 1 \end{vmatrix}}=\frac{\begin{vmatrix} d & d^2 & 1\\ 0 & 0 & 1\\ m-x & (m-x)(-m-x) & 1 \end{vmatrix}}{\begin{vmatrix} d & d & 1\\ 0 & 0 & 1\\ m-x & -m-x & 1 \end{vmatrix}}=\frac{d^2(m-x)-d(m-x)(-m-x)}{d(m-x)-d(-m-x)}=\frac{(m-x)(d+m+x)}{2m}\]\[P\in AB\iff \frac{p-a}{a-b}=\overline{\left(\frac{p-a}{a-b}\right)}\iff \frac{p-a}{a-m+x}=\frac{\overline{p}-a}{a+m+x}\iff \overline{p}=\frac{p(a+m+x)-2am}{a-m+x}\]\[\text{Analogously }P\in CD\iff \overline{p}=\frac{p(d+m-x)-2dm}{d-m-x}\Longrightarrow \frac{p(d+m-x)-2dm}{d-m-x}=\overline{p}=\frac{p(a+m+x)-2am}{a-m+x}\]\[\Longrightarrow p(a+m+x)(d-m-x)-p(a-m+x)(d+m-x)=2am(d-m-x)-2dm(a-m+x)\]\[\Longrightarrow p=\frac{am+ax+dx-dm}{a+2x-d}\]Now $PQ\perp O_{1}O_{2}\iff \frac{p-q}{o_{1}-o_{2}}\in i\mathbb{R}\iff \frac{\frac{am+ax+dx-dm}{a+2x-d}}{\frac{(m+x)(a+m-x)}{2m}-\frac{(m-x)(d+m+x)}{2m}}\in i\mathbb{R}\iff \frac{\frac{am+ax+dx-dm}{a+2x-d}}{\frac{am+ax+dx-dm}{2m}}\in i\mathbb{R}\iff \frac{2m}{a+2x-d}\in i\mathbb{R}$ which is obvious since $m\in i\mathbb{R}$ and $\frac{2}{a+2x-d}\in\mathbb{R}$, so we're done.

A cute problem indeed. Here is a sketch of my solution at the contest: The statement is equivalent to proving that $P$ lies on the radical axis of the circumcircles of triangles $AQC$ and $BQD$. Let their second intersection be point $W$. We will prove that $P, Q$ and $W$ are collinear. Easy anglechase gives us $\angle BWA = \angle CWD$. Let $AC$ and $BD$ intersect at point $O$. From the isogonal lines lemma we get that $WO$ and $WP$ are isogonal conjugates with respect to $\angle CBW$. So it is enough to prove that $WO$ and $WQ$ are isogonal conjugates with respect to the same angle. Now let $AW$ and $BQ$ intersect at point $Y$ and $DW$ and $AC$ intersect at point $X$. Again, easy anglechase gives us that $CXYBW$ is concyclic, as $XY$ and $BC$ are parallel. We now need to find that $O$ lies on this circumcircle, which is easy - just take $O_1$ to be point on $WQ$ such that $OO_1$ is parallel to $BC$ and assume that $O$ and $O_1$ do not lie on the circumcircle of $CXYBW$ - this leads to $AB=CD$ which is a trivial case.

Let $X = (AQC) \cap PD$, $Y= (AQC) \cap PA$ also $M = (BQD) \cap PD$ and $ N = (BQD) \cap PA$. $$\measuredangle BCQ=\measuredangle DQC = \measuredangle AXC = \measuredangle QBC = \measuredangle BQA = \measuredangle BMD$$ hence $AX \parallel BM$ and similarly $ND \parallel YC$. By using power of point $$PY\cdot PA = PX \cdot PC \Rightarrow PY \cdot PB = PM \cdot PC \Rightarrow PA \cdot PY = PD \cdot PM$$Hence $YMAD$ cyclic. which give us $P$ lie on radical axis of both circles.