Let $\Omega$ be the circumcircle of cyclic quadrilateral $ABCD$. Consider such pairs of points $P$, $Q$ of diagonal $AC$ that the rays $BP$ and $BQ$ are symmetric with respect the bisector of angle $B$. Find the locus of circumcenters of triangles $PDQ$.
1. **Define Points and Setup**:
   Let $\Omega$ be the circumcircle of the cyclic quadrilateral $ABCD$. Consider pairs of points $P$ and $Q$ on the diagonal $AC$ such that the rays $BP$ and $BQ$ are symmetric with respect to the bisector of angle $B$. Let $E$ be the intersection of the angle bisector of $\angle B$ with $AC$, and let $F$ be the intersection of line $ED$ with $\Omega$.

2. **Claim**:
   The circumcircle of $\triangle BPQ$ is tangent to $\Omega$.

3. **Proof of Claim**:
   Extend $AQ$ to intersect $\Omega$ at $G$. By spiral similarity, we have:
   \[
   \triangle ABQ \sim \triangle BPC
   \]
   This implies:
   \[
   \angle BQP = \angle EBP
   \]
   This proves that the circumcircle of $\triangle BPQ$ is tangent to $\Omega$.

4. **Cyclic Quadrilateral**:
   Since the circumcircle of $\triangle BPQ$ is tangent to $\Omega$, we have:
   \[
   EP \times EQ = EB^2 = EF \times ED
   \]
   This implies that $PQDF$ is cyclic.

5. **Locus of Circumcenters**:
   Since $PQDF$ is cyclic, the circumcenter of $\triangle PDQ$ moves along the perpendicular bisector of $DF$.

Conclusion:
\[
\boxed{\text{The locus of the circumcenters of triangles } PDQ \text{ is the perpendicular bisector of } DF.}
\]