The data given does not determine the volume of the pyramid.
Note that since $ADM$ and $BCM$ are perpendicular to $ABCD$, it follows that $M$ lies above the intersection $P$ of $AD$ and $BC$. Let $h$ be the height of $M$ above $P$, let $x$ be the perpendicular distance from $P$ to $AB$, and let $d$ be the perpendicular distance between $CD$ and $AB$. Then
$S_{\triangle BCM}=\frac12 h|BC|=9$
and
$S_{\triangle ABM}=\frac12\sqrt{x^2+h^2}|AB|=20\;,$
and with $|AB|=5$ this yields
$x^2+h^2=64\;.$
We also have
$\frac{x-d}x=\frac35\;,$
so $x=\frac52d$, and thus
$\left(\frac52d\right)^2+h^2=64\;.$
Writing $d=|BC|\cos\alpha$, with $\alpha$ the angle between $BC$ and the perpendicular on $AB$ and $CD$, we arrive at
$\left(\frac52|BC|\cos\alpha\right)^2+\left(\frac{18}{|BC|}\right)^2=64\;.$
Solving for $\cos\alpha$ yields
$\cos\alpha=\frac{\sqrt{64-(18/|BC|)^2}}{\frac52|BC|}\;.$
Setting this to zero yields $|BC|\gt\frac94$, and the value is below $1$ in that entire range with a maximum of $32/45$ at $|BC|=\frac94\sqrt2$.
With $V=\frac13hd(|AB|+|CD|)/2=\frac43hd=\frac43h|BC|\cos\alpha=24\cos\alpha$, the volume of the pyramid can take any value between $0$ and $256/15\approx17$.