By symmetry, the midpoint of $P$ and $Q$ is the same as the midpoint of the roots. Use this to find a relationship between $p$ and $q$.

Note that the height of the trapezium can by given by the $y$ coordinate of the point $Q$ (that is, by substituting $x=q$ into the equation of the parabola).

Here are the dimensions of the trapezium

Use the area of the trapezium to show that

Recall that $q\in \mathbb{Z}$. This means that $(2q+1)\in \mathbb{Z}$ and $(2q-3)\in \mathbb{Z}$.

The equation $(2q+1{)}^2(2q-3)=25$ therefore implies that $25$ has been factored into ${\text{odd}}^2\times \text{odd}$. There's only two ways to do that, and only one of them makes sense given the diagram!

The roots of the parabola are $x=-\frac{1}{2}$ and $x=\frac{3}{2}$. The midpoint of the roots is $\frac{1}{2}$. By symmetry, $p$ and $q$ have the same midpoint, so

Now let's consider the area of the trapezium

By the formula for the area of a trapezium, we find that

Note that $h$ is given by the $y$ coordinate of $Q$ (or $P$, but let's focus on getting things in terms of $q$). So $h=(2q+1)(3q-2)$. We also have $p=1-q$. Substituting this information into $(i)$ we get

This implies $25$ has been factored into a square multiplied by an integer. The only ways to do this are $1^2\times 25$ or $5^2\times 1$. We can't have $2q+1=1$ because the diagram implies $q>0$, so we must have

This also implies

and

We conclude that $pqh=-5$.