From the right angle triangle $\triangle PAB$ with the altitude AF, we get $BF \cdot BP = BA^2 = BE^2$, which means that the line BC is a tangent to the circumcircle (O) of the triangle $\triangle EFP$ at the vertex E, i.e., this circumcircle is centerd on the midline of the rectangle ABCD through the midpoint E of BC. Likewise, from the right angle triangle $\triangle PDC$ with the altitude DG, we get $CG \cdot CP = CD^2 = CE^2$, which again means that the line BC is a tangent to the circumcircle (O') of the triangle $\triangle EGP$ at the vertex E, i.e., this circumcircle is also centerd on the midline of the rectangle ABCD through the midpoint E of BC. But since all circles passing through the points E, P are centered on the perpendicular bisector of the segment EP, only one such circle exists, which is also centered on the rectangle midline through E. Thus the circumcircles $(O) \equiv (O')$ of the triangles $\triangle EFP, \triangle EGP$ are identical. In other words, the quadrilateral EFPG is cyclic.

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Yetti, I think that you are not angry cross me ! I continue the your solution differently: from the relations $BE^2=BF\cdot BP,\ CE^2=CG\cdot CP$ results $\triangle BEF\sim \triangle BPE,\ \triangle CEG\sim \triangle CPE\Longrightarrow$ $\widehat {BEF}\equiv \widehat {BPE},\ \widehat {CEG}\equiv \widehat {CPE}\Longrightarrow m(\widehat {FEG})+m(\widehat {FPG})=180^{\circ}.$