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$OA=OB$ is enough to solve the problem (they are both chords of $\gamma$). Best regards, sunken rock

Suppose that $A_0$ is a point on $\gamma$ such that ∠$A_0PC$ = ∠$BPD$. Then the segments $OA_0$ and $OB$ subtends same angle in the respective minor arcs, so $OA_0$ = $OB$. This shows that A lies on $\omega$ and hence $A_0 = A$. This proves that ∠$APC$= ∠$BPD$.

My Solution. Clearly, $OPAB$ is cyclic, as $O,P,A,B\in \gamma.$ Thus, $\angle ABO = 180^{\circ}-\angle OPA=\angle APC.$ And, $\angle BPD=\angle OPB=\angle OAB.$ As $A,B\in \omega,$ and $O$ is the center of $\omega,$ we can say that $OA=OB,$ which implies that $\angle OAB=\angle OBA.$ Now, we can say that $\angle BPD=\angle OAB=\angle ABO=\angle APC\Rightarrow \boxed{\angle APC=\angle BPD}~.~\Square$

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Inverting the problem through $\omega$ directly solves it

Let $\angle OBA=\angle OAB=\angle OPB=x$ $\implies$ $\angle BOA=\angle BPA=180^{\circ}-2x$ $\implies$ $\angle OPB=\angle DPB=x=\angle APC$

cute(ugly) solution : invert the diagram with P and denote A becoming A (for we don't need old points) . since the inversion don't change ∠(omega,line CD) ,CD is still the diameter of the reflection of omega . since DCPO are harmonic and ∠DBC is 90°(∵diameter CD), ∠ABD=∠PBD . On the other hand ,∠ABD=∠BCD, so ∠PBD=∠BCD ,showing the intersection of BD and AC point G and P,C,B lie on the same circle . In the same way we know G,P,A,D lie on the same circle , so ∠APD=∠AGD=∠BGC=∠BPC ,which is equivalent to the original problem.