Irish Mathematical Olympiad, 2014

Join $P$ to $E$, $F$, $G$ and $H$. The details are worked out below for the situation in the picture provided where $E$ and $H$ are both between $B$ and $C$, $F$ is between $A$ and $D$ but $G$ is not between $A$ and $D$. For other possible positions of $E$, $F$, $G$, $H$, after replacing some angles with their supplements, the arguments provided will also work.
Alternatively, one could avoid considering several cases by working with oriented angles modulo $180^\circ$, see for example Section 1.7 in [1].

From the cyclic quadrilateral $AGHP$ we get $\angle PHG = \angle PAD = \angle CAD$. Because $A$, $B$, $C$, $D$ are concyclic we obtain $\angle CAD = \angle CBD$. As $H$, $B$, $G$, $P$ are on a circle, $\angle CBD = \angle HBP = \angle HGP$. Hence $\angle PHG = \angle HGP$ and so $|PG| = |PH|$.

Similarly, using twice that $P$, $E$, $C$, $D$, $F$ are on a circle and that $ABCD$ is cyclic, we obtain
$$\angle PEF = \angle PDF = \angle BDA = \angle BCA = \angle ECP = \angle EFP$$
and this implies $|PE| = |PF|$.

Using all three circles, we get the following equalities
$$\angle PEH = \angle CDP = \angle CDB = \angle CAB = \angle PAB = \angle PHE$$
from which we get $|PE| = |PH|$. Altogether we have shown that $E$, $F$, $G$, $H$ all have the same distance from $P$.