National Olympiad of Argentina

Extend $PA$ and $PB$ to meet $CD$ at $X$ and $Y$ respectively. Since $XY \parallel AB$, Thales' theorem yields $\frac{XD}{AE} = \frac{PD}{PE} = \frac{CD}{FE}$. Also $AE = FE$, so $XD = CD$, i.e. $D$ is the midpoint of $XC$. In addition $CD = DA$, hence $DA = DX = DC$. Thus triangle $XCA$ is right at $A$, so that $PA \perp AC$. By symmetry $PB \perp BD$. Let $O$ be the circumcenter of $ABCD$ (an isosceles trapezoid is cyclic). Then $O$ and $D$ are both equidistant from $A$ and $C$, hence $OD$ is the perpendicular bisector of segment $AC$. In particular $OD \perp AC$ and likewise $OC \perp BD$.

Now $PA \perp AC$, $PB \perp BD$ yield $PA \parallel OD$; likewise $PB \parallel OC$. Hence $\angle APB = \angle DOC$. Note that $A$ and $O$ are on the same side of chord $CD$ in the circumcircle, hence $\angle DOC = 2\angle DAC$. Note also that $AC$ is the bisector of $\angle DAB$ because $CD = CB$ due to $CD = CB$. It follows that $\angle DAB = 2\angle DAC = \angle DOC$ and $\angle APB = \angle DOC = \angle DAB$.