jmc

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: $$\begin{array}{rl}s& =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & =\frac{4\pm \sqrt{4^2-4(-49)(1)}}{2(1)}\\ & =\frac{4\pm \sqrt{4(4+49)}}{2}\\ & =2\pm \sqrt{53}.\end{array}$$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: $$\frac{1}{p^2}+\frac{1}{q^2}=\frac{1}{p^2}\cdot \frac{q^2}{q^2}+\frac{1}{q^2}\cdot \frac{p^2}{p^2}=\frac{q^2+p^2}{p^2{q}^2}.$$We know from above that $p^2+q^2=s^2-2pq=s^2-2s$, and $p^2{q}^2=(pq{)}^2=s^2$, so we must find $$\begin{array}{rl}\frac{1}{\frac{s^2-2s}{s^2}}& =\frac{s^2}{s^2-2s}\\ & =\frac{s}{s-2}\\ & =\frac{2+\sqrt{53}}{2+\sqrt{53}-2}\\ & =\frac{2+\sqrt{53}}{\sqrt{53}}.\end{array}$$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 $$\begin{array}{rl}a+b+c+d& =53+2+53+53\\ & =\boxed{161}.\end{array}$$