Rectangle $ABCD$ is the base of pyramid $PABCD$. If $AB = 3$, $BC = 2$, $\overline{PA}\perp \overline{AD}$, $\overline{PA}\perp \overline{AB

This is a funny lookin pyramid! The formula for the volume of this pyramid is the same as if "P" were above the center of the base:  volume = (1/3)(area of base)(height)

This helped me verify that.

Draw AC.

Now look at triangle ABC.  From the Pythagorean theorem....

AC²   =   2² + 3²   =   4 + 9   =   13

AC is in the same plane as AD and AB, so PA is perpendicular to AC.

Look at triangle PAC.  From the Pythagorean theorem again....

AC² + PA²  =  5²

13  +  PA²  =  25

PA²  =  12

PA  =  √12

PA  =  2√3

And....

volume of pyramid  = (1/3)(area of base)(height)

                               =  (1/3)(  2 * 3  )( PA )

                               =  (1/3)(6)(2√3)

                               =  4√3     cubic units

*edit*

I just remembered how my math teacher explained that the volume of an oblique cylinder is the same as the volume of a right cylinder (with the same height and base).

Imagine a stack of pennies. If you line the stack straight up, it is a right cylinder. And if you lean the stack over, it is oblique. But the height, area of the base, and the volume stays the same.

And in this case, imagine a stack of square pieces of cardboard that get smaller as you go higher. You can stack the squares up to form a right pyramid, or you can slide them to look like the pyramid in this problem. The volume stays the same.