Inspired by Jake Lai

If we first consider the product over all primes (call it $Q$ ) then we have by using the Euler product:

$\large Q=\prod_{\text{Primes}} \frac{p^4+1}{p^4-1}=\prod_{\text{Primes}} \frac{1+\frac{1}{p^4}}{1-\frac{1}{p^4}}=\prod_{\text{Primes}} \frac{1-\frac{1}{p^8}}{\left(1-\frac{1}{p^4}\right)^2}=\frac{\zeta(4)^2}{\zeta(8)}$

You can work out the exact values of $\zeta(2n)$ for all non-negative integers $n$ with the Bernoulli numbers.

I have the first few values memorised and it comes out as $\large Q=\frac{\left(\frac{\pi^4}{90}\right)^2}{\frac{\pi^8}{9450}}=\frac{7}{6}$ .

It's easy enough to see that $\large Q=\frac{2^4+1}{2^4-1} P$ since 2 is the only even prime.

We simplify and get $\large P=\frac{35}{34}$ .