The key idea for this problem lies in the following lemma:

Lemma: Let $ABC$ be a triangle and $X$ a point in the interior of angle $\angle BAC$ such that $\angle ABX = \angle ACX$. Define $Y$ such that $BXC$ is a parallelogram. Then $AX$ and $AY$ are isogonal with respect to $\angle BAC$.

Proof. Consider the triangle $ABC$. Since $BXC$ is a parallelogram, we have $BX \parallel AC$ and $CX \parallel AB$. By the alternate interior angles theorem, we have $\angle XBA = \angle XCB$. Since $\angle ABX = \angle ACX$, we can conclude that $\angle XAB = \angle XCA$. Thus, $AX$ and $AY$ are isogonal with respect to $\angle BAC$. $\square$

Now, let's proceed with the solution to the main problem. We need to prove that the circle $(POQ)$ passes through a fixed point as $A, B, C, D$ vary according to the given restrictions.

Let $O'$ be the intersection of lines $BD$ and $AC$. Since $AC \parallel XT$ and $BD \parallel XY$, by the Lemma, we know that $AP$ and $AQ$ are isogonal with respect to $\angle XO'Y$.

Since $ABCD$ is a cyclic quadrilateral, we have $\angle ABC = \angle ADC$. Thus, $\angle PBC = \angle PDC$. This implies that $PB$ and $PD$ are isogonal with respect to $\angle ABC$. Similarly, $QA$ and $QC$ are isogonal with respect to $\angle ADC$.

Therefore, we have $\angle ABP = \angle DCQ$ and $\angle BAP = \angle CQD$. Combining these equalities, we get $\angle ABO' = \angle DCO'$. This implies that $ABO'D$ is a cyclic quadrilateral.

Let $O$ be the circumcenter of $ABO'D$. Since $ABCD$ is a cyclic quadrilateral, $O$ is also the circumcenter of $ABCD$. Therefore, $O$ lies on the circle $(POQ)$.

Thus, we have shown that as $A, B, C, D$ vary according to the given restrictions, the circle $(POQ)$ passes through the fixed point $O$. This completes the proof.