I am manually trying to compute prime-zeta of 2. Apologies if those $p\binom{}{}$ terms appear wierd.
It is a convention I have adopted.
$p\binom{2}{x} = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots (x \hspace{3pt} \text{terms})$
If $x \rightarrow \infty$, we get the prime-zeta of 2, which is probably represented as, $P(2)$.
Its value is already known.
I started with $\zeta(2)$, rearranged it for form a series of the form $\sum_p\sum_{n=0}^\infty \frac{1}{(p^{2^n})^2}$ (i.e. the denominator is $p^{2^n}$) and proceeded to fill the gaps in between the two consecutive terms in each of the series. At some point, that series appears, along with some terms in $P(4)$. And because I saw the latter two, I wonder if its worth proceeding with more simplifications.