Answer : (a) ∠CPA = ∠DPB
In the bigger circle, ∠APX = ∠ABP
In the smaller circle, ∠CPX = ∠PDC {Angles in alternate segment are equal.}
⇒ ∠APX + ∠CPA = ∠CPX = ∠PDC
⇒ ∠ABP + ∠CPA = ∠PDC (∵ ∠APX = ∠ABP)
⇒ ∠ABP + ∠CPA = ∠DBP + ∠DPB (ext. ∠ theorem in ∆ PDB)
∠ABP + ∠DPB
⇒ ∠CPA = ∠DPB.