Is there a simple way?

The term $\frac{1}{p^2}\prod_{q <p} \frac{q^2-1}{q^2}$ is equal to the sum of all terms of the form $\frac{(-1)^{k-1}}{p_1^2p_2^2 \cdots p_k^2}$ where $p_1,p_2,...,p_k$ are prime and $p_1 < p_2 < \cdots < p_k=p$ . Thus the required sum is equal to $\sum_p \frac{1}{p^2}\prod_{q < p}\frac{q^2-1}{q^2} \; = \; 1 - \prod_p \left(1 - \frac{1}{p^2}\right) \; = \; 1 - \zeta(2)^{-1} \; = \; 1 - \frac{6}{\pi^2}$ making the answer $1 + 6 + 1 + 2 = \boxed{10}$ .

I will denote by $p_n$ the $n^{th}$ prime number, $m\in\mathbb{N}^*$ .

$\begin{aligned} \displaystyle\sum_{n=1}^{N} \frac{1}{p_n^m} \displaystyle\prod_{k<n}(1-\frac{1}{p_k^m}) &= -\displaystyle\sum_{n=1}^{N} (1-\frac{1}{p_n^m}) \displaystyle\prod_{k<n}(1-\frac{1}{p_k^m}) + \displaystyle\sum_{n=1}^{N} \displaystyle\prod_{k<n}(1-\frac{1}{p_k^m}) \\ & = -\displaystyle\sum_{n=2}^{N+1} \displaystyle\prod_{k<n}(1-\frac{1}{p_k^m}) +\displaystyle\sum_{n=1}^{N} \displaystyle\prod_{k<n}(1-\frac{1}{p_k^m}) \\ & = \displaystyle\prod_{k<1}(1-\frac{1}{p_k^m})-\displaystyle\prod_{k<N+1}(1-\frac{1}{p_k^m})\\ & = 1-\displaystyle\prod_{k<N+1}(1-\frac{1}{p_k^m}) \end{aligned}$

Therefore,

$\displaystyle\sum_{n=1}^{N} \frac{1}{p_n^m} \displaystyle\prod_{k<n}(1-\frac{1}{p_k^m})\underset{N\to +\infty}{\longrightarrow} 1-\frac{1}{\zeta(m)}\\ \text{ie}, \ \boxed{\displaystyle\sum_{n=1}^{+\infty} \frac{1}{p_n^m} \displaystyle\prod_{k<n}(1-\frac{1}{p_k^m}) = 1-\frac{1}{\zeta(m)}}$

After breaking the summation we easily can note a very familiar series. $\sum_{primes} \frac{1}{p^2} \prod_{q <p} (\frac{q^2-1}{q^2})$

$=\frac{1}{2^2} +\frac{1}{3^2} - \frac{1}{{2^2}{3^2}} - \frac{1}{5^2} -\frac{1}{{2^2}{5^2}} -\frac{1}{{3^2}{5^2}} +\frac{1}{{2^2}{3^2}{5^2}} -....=$

$\sum_{p} \frac{1}{p^2} - \sum_{p ,q} \frac{1}{{p^2}{q^2}}+\sum_{p,q,r} \frac{1}{{p^2}{q^2}{r^2}}-....$ .

And this equals to

$1-\prod_{primes} (1-\frac{1}{p^2}) =1- \frac{1}{\zeta(2)}=1-\frac{6}{π^2}$

So, $a=1,b=6, c=1, d=2$ .

So, $a+b+c+d=10$ .

[ $p,q,r....$ denote primes].