1. Identify the given elements and their properties:
- Let $\omega$ be a circle with center $O$.
- Let $\gamma$ be another circle passing through $O$ and intersecting $\omega$ at points $A$ and $B$.
- The diameter $CD$ of $\omega$ intersects $\gamma$ at a point $P$ different from $O$.

2. **Establish that $OPAB$ is a cyclic quadrilateral:**
- Since $O$, $P$, $A$, and $B$ all lie on the circle $\gamma$, the quadrilateral $OPAB$ is cyclic.

3. Use the properties of cyclic quadrilaterals:
- In a cyclic quadrilateral, opposite angles sum to $180^\circ$. Therefore, $\angle OPA + \angle OBA = 180^\circ$.

4. **Relate the angles $\angle APC$ and $\angle BPD$:**
- Since $O$ is the center of $\omega$, $OA = OB$ (radii of the circle $\omega$).
- This implies that $\triangle OAB$ is isosceles with $OA = OB$.
- Therefore, $\angle OAB = \angle OBA$.

5. **Express $\angle APC$ and $\angle BPD$ in terms of the angles in $\triangle OAB$:**
- Since $OPAB$ is cyclic, $\angle OPA = \angle OBA$ (opposite angles in a cyclic quadrilateral).
- Similarly, $\angle OPB = \angle OAB$.

6. Conclude the equality of the angles:
- From the above, we have $\angle APC = \angle OPA$ and $\angle BPD = \angle OPB$.
- Since $\angle OPA = \angle OBA$ and $\angle OPB = \angle OAB$, and $\angle OAB = \angle OBA$, it follows that $\angle APC = \angle BPD$.

Therefore, we have shown that $\angle APC = \angle BPD$.

$\blacksquare$