since AQ is a diameter, you have angle ACQ = 90 degrees

also, angle APQ = 90 degrees

let angles AQC = DQB = a (vertically opposite angles)

so you have angle CAQ =

since APD is straight line,

angles DPQ = APQ = 90 degrees

so you can prove that QD is diameter of circle PQBD

Therefore, angle QBD = 90 degrees

now you can prove that triangles ACQ and DBQ are similar,

with angles CAQ = BDQ =

Hence, ADBC is an inscribed quadrilateral, and so angles ADC = ABC