is this how this is solved?
No. Since $\displaystyle \Delta ADP\,\cong \Delta BFP$ then $\displaystyle AP\,=\,PB$.
A bisector divides a segment into two equal lengths and since they are already equal, then $\displaystyle PE\,=\,KP$
EDIT: Your title says prove a congruent angle but your file says prove a side is the same. Is that what you intended?
We can call $\displaystyle \measuredangle~ADK=\alpha\implies\measuredangle~KD P=\measuredangle~EFP=\alpha$, and $\displaystyle \measuredangle~KPD=\measuredangle~EPF\implies$ by ASA congruence we have $\displaystyle \triangle{DKP}\cong\triangle{FEP}\implies\overline {PE}=\overline{KP}\,\blacksquare$