In rectangle $ABCD$, points $E$ and $F$ lie on segments $AB$ and $CD$, respectively, such that $AE = \frac{AB}{8}$ and $CF = \frac{CD}{5}$. Segment $BD$ intersects segment $EF$ at $P$. What fraction of the area of rectangle $ABCD$ lies in triangle $EBP$? Express your answer as a common fraction.