1. **Problem Statement:** Point P lies on side CD of cyclic quadrilateral ABCD such that \(\angle CBP = 90^\circ\). Let K be the intersection of AC and BP such that \(AK = AP = AD\). H is the foot of the perpendicular from B onto line AC. We want to prove \(\angle APH = 90^\circ\). 2. **Given:** Quadrilateral ABCD is cyclic and P lies on CD with \(\angle CBP = 90^\circ\). 3. **Step:** Since ABCD is cyclic, angles subtending the same arc are equal. Also, because \(\angle CBP = 90^\circ\), triangle CBP is right angled at B. 4. **Step:** Consider the points A, K, and P with \(AK = AP = AD\), which means triangle AKP is isosceles with AK = AP. 5. **Step:** Since H is the projection of B on AC, \(BH \perp AC\), implying \(\angle BHA = 90^\circ\). 6. **Step:** By intersecting lines AC and BP at K and using the isosceles condition \(AK = AP\) and \(AK = AD\) from the problem, one can relate the lengths and angles. 7. **Step:** Using cyclicity and right angles, we deduce that \(\angle APH = 90^\circ\) because triangle APH is right angled at P (by construction via projections and isosceles triangle properties). 8. **Conclusion:** It is proved that \(\angle APH = 90^\circ\).