the attachment shows a general case.
alternate segment theorem that the triangles PDB and ADP are similar. Deduce that . (In fact, this is a well-known theorem: from a point D, draw a tangent DP to a circle, and a chord DBA cutting the circle; then the square of PD is equal to the product of DA and DB.)
Now do the same for the other circle, to see that is also equal to . Conclusion: PD = QD.