Answer: A

Small Hint:

Let the side lengths be $L$ and $W$, and write equations for $L+W$ and $L^2+W^2$

Big Hint:

Write $2(L^2+W^2)$, then subtract $(L+W)^2$ to obtain $(L-W)^2$

Solution:

The perimeter and diagonal give $$\begin{aligned} L+W&=\frac p2,\\ L^2+W^2&=d^2. \end{aligned}$$ Hence $$\begin{aligned} (L-W)^2 &=2(L^2+W^2)\\ &\quad{}-(L+W)^2\\ &=2d^2-\frac{p^2}{4}\\ &=\frac{8d^2-p^2}{4}. \end{aligned}$$ Taking the nonnegative square root gives $$L-W=\frac{\sqrt{8d^2-p^2}}2.$$

Thus, the correct answer is A.