Extend FP intersection AD to point g.
Ideas:
1) according to the square, we get: AC bisecting angle BAD,
So angle BAC= angle DAC=45 degrees.
2) provable: PE=PG, PF=GD.
Because: EPF angle = PGD angle =90 degrees.
So: triangle EPF is equal to triangle PGD.
So: EF=DP