First note that $\angle BAC = \angle BDC$ (inscribed angles) and $\angle BAC = \angle ACD$ (alternate angles at $AB \parallel CD$). Hence, $\angle BDC = \angle ACD$ and triangle $CDP$ is isosceles with $|CP| = |DP|$. This implies that $P$ is on the perpendicular bisector of $CD$ and that $\angle PBA = \angle BAP = 45^\circ$ (because $AC \perp BD$).
The centre $O$ of the circle is on the perpendicular bisectors of the chords $AB$ and $CD$. As $AB$ and $CD$ are parallel, their perpendicular bisectors are parallel as well. As they have $O$ in common, they coincide. This means that $OP$ is the common perpendicular bisector of $AB$ and of $CD$.

Because $\angle BEP = \angle CPD = 90^\circ$ and $\angle PDC = \angle PBA = 45^\circ$, triangles BEP and CPD are right angled and isosceles. In particular, $|BE| = |EP|$. Moreover, $\angle BOC = 2\angle BDC = 90^\circ$ (central angle), hence
$$\angle BOE = 180^\circ - \angle BOC - \angle COF = 90^\circ - \angle COF = \angle OCF.$$
Since $|OB| = |OC|$, we can now use ASA to see that $\triangle OCF$ is congruent to $\triangle BOE$, thus $|OF| = |BE| = |EP|$.