To find the volume of a pyramid, I need to multiply the area of the base by the height and divide by 3. The base is a rectangle, so its area is just the product of its length and width, which are 8 and 4, respectively. So the area of the base is 8 times 4, which is 32. The height of the pyramid is the perpendicular distance from the vertex $P$ to the base. Since $\overline{PA}$ is perpendicular to two adjacent edges of the base, it must also be perpendicular to the plane containing the base, and therefore it is the height of the pyramid. So I just need to find the length of $\overline{PA}$. I notice that triangle $PAB$ is a right triangle, so I can use the Pythagorean theorem to find the length of $\overline{PA}$. By the Pythagorean theorem, $PA^2 + AB^2 = PB^2$. Plugging in the values given in the problem, I get $PA^2 + 8^2 = 17^2$. Simplifying, I get $PA^2 + 64 = 289$. Subtracting 64 from both sides, I get $PA^2 = 225$. Taking the square root of both sides, I get $PA = 15$. So the height of the pyramid is 15. Now I can find the volume by multiplying the area of the base by the height and dividing by 3. The volume is 32 times 15 divided by 3, which is 160. 
Rectangle $ABCD$ is the base of pyramid $PABCD$. If $AB = 8$, $BC = 4$, $\overline{PA}\perp \overline{AD}$, $\overline{PA}\perp \overline{AB}$, and $PB = 17$, then what is the volume of $PABCD$?