Let $ p+q=pq=s $. Then $ (p+q)^2=p^2+q^2+2pq=s^2 $. We subtract $ 4pq=4s $ from both sides to find $$p^2+q^2-2pq=(p-q)^2=s^2-4s.$$We are given that the difference between $ p $ and $ q $ is $ 7 $, so $ p-q=\pm 7 $, and $ (p-q)^2=(\pm 7)^2=49 $, so our equation becomes $ 49=s^2-4s $ or $ s^2-4s-49=0 $. We can solve for $ s $ using the quadratic formula: \begin{align*} s&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\frac{4\pm\sqrt{4^2-4(-49)(1)}}{2(1)}\\ &=\frac{4\pm\sqrt{4(4+49)}}{2}\\ &=2\pm\sqrt{53}.\end{align*}Since $ p $ and $ q $ are positive, we know $ s=pq=p+q $ is positive, so we take the positive solution, $ s=2+\sqrt{53} $. Now we must find $ \frac{1}{\frac{1}{p^2}+\frac{1}{q^2}} $. We can combine the fractions in the denominator by finding a common denominator: $$\frac{1}{p^2}+\frac{1}{q^2}=\frac{1}{p^2}\cdot\frac{q^2}{q^2}+\frac{1}{q^2}\cdot\frac{p^2}{p^2}=\frac{q^2+p^2}{p^2q^2}.$$We know from above that $ p^2+q^2=s^2-2pq=s^2-2s $, and $ p^2q^2=(pq)^2=s^2 $, so we must find \begin{align*} \frac{1}{\frac{s^2-2s}{s^2}}&=\frac{s^2}{s^2-2s}\\ &=\frac{s}{s-2}\\ &=\frac{2+\sqrt{53}}{2+\sqrt{53}-2}\\ &=\frac{2+\sqrt{53}}{\sqrt{53}}.\end{align*}Rationalizing the denominator gives $ \boxed{\frac{2\sqrt{53}+53}{53}} $. Thus in the form requested, $ a=53 $, $ b=2 $, $ c=53 $, and $ d=53 $, so \begin{align*} a+b+c+d&=53+2+53+53\\ &=\boxed{161}.\end{align*}