Final Round of the 73rd Czech and Slovak Mathematical Olympiad (March 17–20, 2024)

From the given equalities, one sees that $PC\parallel AD$ and $PD\parallel BC$. Since the lines $PC$ and $PD$ are distinct (we know that $\angle CPD\ne 0$), the lines $AD$ and $BC$ are not parallel, so they intersect at a unique point $X$ such that $PCXD$ is a rhombus.

Now, note that the quadrilateral $AXCP$ is an isosceles trapezoid, since $AX\parallel CP$ and $\angle PAX=\angle CXA$. Therefore, the perpendicular bisectors of its bases $AX$ and $CP$ coincide. Since $O$ is the circumcentre of $CPD$, it lies on the bisector of $CP$, hence also on the bisector of $AX$, and so we have $OX=OA$. By considering the isosceles trapezoid $BXDP$, we can also obtain $OX=OB$, which gives us the desired equality.