Find the area of the quadrilateral ABCD
given $EA=BE$ and $BP=PC$ and $PE\perp BD$
$EA=BE$ and $BP=PC$ $\implies \frac{BE}{BP}=\frac{EA}{PC} \implies PE\parallel AC$
$\implies AC\perp BD$
$\implies $ Area of the quadrilateral ABCD=Area of the rectangle IJKL
But, for that we need the other diagonal AC, is there any way to find it ?
In triangle $ABC$, since $E$ is the middle of $[AB]$, and $P$ is the middle of $[BC]$, this implies that $EP$ is a midsegment in triangle $ABC$, therefore it is parallel to $AC$ (as you mentioned), but also $$EP = \frac{AC}{2} \Rightarrow AC=2EP=10$$
Since $PE ∥AC$ and $PE⊥BD$, we have $AC⊥BD$, so $ABCD$ is a orthodiagonal quadrilateral, hence its area is $$A_{ABCD} = \frac{AC \cdot BD}{2} = PE \cdot BD = 65$$
P.S. The formula for the midsegment is obvious from Thales's Theorem.