Since $ABCD$, $CDEF$ and $EFGH$ are squares, then we have $AD:DE:EH = 1: 1:1$ and $BC: CF:FG=1:1:1$.
Since $\Delta ADP \sim \Delta AEQ$, then we have
$\begin{array}{rcl}
\dfrac{\text{Area of $\Delta ADP$}}{\text{Area of $\Delta AEQ$}} & = & \left( \dfrac{AD}{AE} \right)^2 \\
\dfrac{\text{Area of $\Delta ADP$}}{\text{Area of $\Delta AEQ$}} & = & \left( \dfrac{1}{2} \right)^2 \\
\dfrac{\text{Area of $\Delta ADP$}}{\text{Area of $\Delta AEQ$}} & = & \dfrac{1}{4}
\end{array}$
Hence, we have
$\begin{array}{rcl}
\dfrac{\text{Area of $\Delta ADP$}}{\text{Area of quadrilateral $DEQP$}} & = & \dfrac{1}{3} \\
\end{array}$
Since $\Delta ADP \sim \Delta AHG$, then we have
$\begin{array}{rcl}
\dfrac{\text{Area of $\Delta ADP$}}{\text{Area of $\Delta AHG$}} & = & \left( \dfrac{AD}{AH} \right)^2 \\
\dfrac{\text{Area of $\Delta ADP$}}{\text{Area of $\Delta AHG$}} & = & \left( \dfrac{1}{3} \right)^2 \\
\dfrac{\text{Area of $\Delta ADP$}}{\text{Area of $\Delta AHG$}} & = & \dfrac{1}{9}
\end{array}$
Hence, we have
$\begin{array}{rcl}
\dfrac{\text{Area of $\Delta ADP$}}{\text{Area of quadrilateral $DHGP$}} & = & \dfrac{1}{8} \\
\end{array}$
Therefore, we have
$\begin{array}{rcl}
\dfrac{\text{Area of quadrilateral $DEQP$}}{\text{Area of quadrilateral $DHGP$}} & = & \dfrac{3}{8} \\
\dfrac{\text{Area of quadrilateral $DEQP$}}{\text{Area of quadrilateral $EHGQ$}} & = & \dfrac{3}{5} \\
\end{array}$
Since the area of quadrilateral $ABCP$ $=$ the area of quadrilateral $EHGQ$, then we have
$\begin{array}{rcl}
\dfrac{\text{Area of quadrilateral $DEQP$}}{\text{Area of quadrilateral $ABCP$}} & = & \dfrac{3}{5} \\
\end{array}$
