$\frac{5}{12}$

Let $AB$ and $AD$ be of length $b$ and $h$ respectively. Then
the area of $ABCD = bh$ and the area of $$ABQP =
\frac{1}{2}b\left(\frac{h}{2} + \frac{h}{3}\right)=
\frac{5bh}{12}$$

Thus the required ratio is $\frac{5}{12}$.

*This problem is taken from the UKMT Mathematical Challenges.*