Answer: 594

Small Hint:

After $AP$ and $BP$ are joined, vertices $A$ and $B$ become one vertex; determine all six edge lengths of the tetrahedron

Big Hint:

Place $C,$ $D,$ and the joined vertex in one coordinate plane, then locate $P$ from its equal distances to the other vertices

Solution:

After folding, $A$ and $B$ become one vertex $X.$ Each half-diagonal of the rectangle has length $\frac{\sqrt{939}}{2}.$ Thus $$XP=CP=DP=\frac{\sqrt{939}}2,$$ while $XC=XD=13\sqrt3$ and $CD=12\sqrt3.$

Place $$\begin{aligned}C&=(-6\sqrt3,0,0),\\D&=(6\sqrt3,0,0),\end{aligned}$$ and $X=(0,\sqrt{399},0).$ These coordinates give the required lengths from $X$ to $C$ and $D.$ Because $P$ is equidistant from $C$ and $D,$ write $P=(0,u,h).$ Equating $PC^2$ and $PX^2$ gives $$u=\frac{291}{2\sqrt{399}}.$$ Then $PC^2=\frac{939}{4}$ yields $$h^2=\frac{507}{4}-u^2=\frac{9801}{133},$$ so $h=\frac{99}{\sqrt{133}}.$

The base triangle $XCD$ has area $$\frac12(12\sqrt3)(\sqrt{399})=18\sqrt{133}.$$ Therefore the pyramid’s volume is $$\frac13(18\sqrt{133})\left(\frac{99}{\sqrt{133}}\right)=594.$$