1. Given that $BCTP$ is cyclic, we know that $\angle BCP = \angle BTP$ and $\angle CBP = \angle CTP$.

2. Since $AP = PT = TD$ and $QB = BC = CR$, triangles $\triangle QBC$ and $\triangle PTD$ are isosceles.

3. Because $BCTP$ is cyclic, we have $\angle QBC = \angle PTD$. This follows from the fact that $\angle QBC$ and $\angle PTD$ are both subtended by the same arc in their respective circles.

4. Since $\triangle QBC$ and $\triangle PTD$ are isosceles, we have $\angle QCB = \angle QBC$ and $\angle PTD = \angle PDT$.

5. Therefore, $\angle BAT = \angle BRT$. This implies that quadrilateral $ABTR$ is cyclic because opposite angles of a cyclic quadrilateral are supplementary.

6. Similarly, we can show that $QPCD$ is cyclic. Since $QPCD$ is cyclic, $\angle QPC = \angle QDC$.

7. Let $CP \cap BR = K$. We need to show that $\angle ARB = \angle ATB$.

8. Since $BCTP$ is cyclic, $\angle ATB = \angle ATP + \angle PTB = \angle RBC + \angle PCB$.

9. Since $AR || PC$, we have $\angle ARB = \angle RKC$.

10. Similarly, we can show that $QD || BT$. Therefore, $AQRD$ is cyclic.

Conclusion:
Since we have shown that $AQRD$ is cyclic, the proof is complete. $\blacksquare$