My solution: Let $O_1,O_2$ be the circumcenter of $(ADP)$ and $(BQD)$. $PQ$ cuts $(ADP)$ at $R$. We have $\angle RAD=\angle RPD=\angle BPQ$, and from $PQ \perp AC$, $\Longrightarrow$ $\angle ADR=90^o$. And $\angle BDR=180^o-\angle RAC=180^o-\angle BPQ-\angle DAC=\angle BQR$ $\Longrightarrow$ $R$ lies on $(BQD)$ $\Longrightarrow$ $O_1O_2\perp RD$ $\Longrightarrow$ the line passes through the circumcenters of triangles $APD$ and $BQD$ is parallel to $AD$.

Let the perpendicular from AD at D intersect the circumcircle of APD at X. Then, AX is a diameter of the circle, which implies $XP\perp AP$, so X lies on $PQ$. Now, $\angle DBP=\angle DAP=\angle DXP$, so DXPB is cyclic. Now we know that DX, which is the radical axis of the two desired circles, is perpendicular to AD, which means the line connecting the center of the two circles is parallel to AD.

Solved with Alex Zhao, Ankit Bisain, Jeffrey Chen, Kevin Wu, Reagan Choi, and Rey Li. Let \(R\) be the anitpode of \(A\) on \((APD)\), so \(Q\), \(P\), \(R\) collinear. Then \(\measuredangle ARQ=\measuredangle ADP=\measuredangle ACQ\), so \(ARCQ\) is cyclic. At last \(PQ\cdot PR=PA\cdot PC=PB\cdot PD\), the end.

Let $O_{1},O_{2}$ be centers of $(APD),(BQD)$, and $T=(APD)\cap (BQD)\backslash D$. Since $\measuredangle PAD=\measuredangle QBD,$ spiral similarity $\left(D;(APD)\mapsto (BQD)\right)$ maps $P$ onto $Q$, so $T\in PQ$. Finally $\angle (AD;DT)=\angle (AC;QP)=\frac{\pi}{2}=\angle (O_{1}O_{2};DT)$ and $AD\parallel O_{1}O_{2}.$