Let $F$ be the mid-point if $AB$, then $P$, $E$, $O$, $F$ are collinear. We have $\angle AOB = 2\angle ACB$, hence using symmetry, $\angle FOB = \angle ACB$. This implies that $CEOB$ is a cyclic quadrilateral, which implies
$$\angle ECO = \angle EBO.$$
From $\triangle EOB$ we have $\angle EBO = \angle FOB - \angle BEO$. By symmetry, $\angle BEO = \angle DEP = \angle CEP$, hence $\angle EBO = \angle FOB - \angle CEP = \angle ACB - \angle CEP = \angle OPC$. This implies $\angle ECO = \angle OPC$, hence triangles $POC$ and $COE$ are similar and so, using $|OC| = |OB|,$
$$\frac{|OE|}{|OB|} = \frac{|OB|}{|OP|}$$
which gives $|OE| \cdot |OP| = |OB|^2 = R^2$.

From $\triangle BOD$ we see that $\angle OBE = (180^\circ - \angle BOD)/2$. The central angle $\angle BOD$ stands on the same arc as $\angle BAD$, hence $\angle BOD = 2\angle BAD = 2\angle ABC$, by symmetry. Finally, because $PO$ is perpendicular to $AB$, we have $90^\circ - \angle ABC = \angle OPB$. Altogether we obtain $\angle OBE = \angle OPB$. As $\triangle OBE$ and $\triangle OPB$ share an angle at $O$, it follows now that these two triangles are similar. This implies
$$\frac{|OE|}{|OB|} = \frac{|OB|}{|OP|}$$
which gives $|OE| \cdot |OP| = |OB|^2 = R^2$.

Considering the external angle of $\triangle AEB$ at $E$ we see that $\angle CEB = \angle CAB + \angle DBA = 2\angle CAB$, because of symmetry. The Central Angle Theorem gives now $\angle COB = 2\angle CAB = \angle CEB$, hence $CEOB$ is a cyclic quadrilateral. By reflection symmetry, $DEOA$ is cyclic as well.
Inversion in the circle through $ABCD$ maps the line $PB$ to the circumcircle of $COB$ and the line $PA$ to the circumcircle of $DOA$. Hence, the point $P$ is inverted to the intersection point (different from $O$) of these two circles, which is $E$. This means that $|OE| \cdot |OP| = R^2$.