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By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

Similarly,

∠BOC = 2∠BAC

Adding above equations, we get,

⇒ ∠AOD+∠BOC = 2(∠BAC + ∠DCA) ----------- (1)

In ΔAPC, ∠PAC + ∠PCA = ∠BPC by exterior angles property.

⇒ ∠BAC + ∠DCA = ∠BPC

Hence, proved.

If AOD and BOC are supplementary to each other,

⇒ ∠BAC + ∠DCA = 90

And from above theorem, ∠BPC = 90°