We have $$\sum_k\frac{(-1)^kP(nk)}{k}=\sum_{k,p}\frac{(-1)^k}{kp^{nk}}=-\sum_p\ln\left(1+\frac{1}{p^n}\right)=\sum_p\ln\left(\frac{1-\frac{1}{p^{2n}}}{1-\frac{1}{p^n}}\right)=\ln\frac{\zeta(n)}{\zeta(2n)}. $$

I assume that in your formula you meant $(-1)^{k-1}$ not $(-1)^k$. If so, this computation shows that your guess is correct whenever the numerator of $\zeta(2n)/\pi^{2n}$ is equal to $1$ (more or less for $n$ up to $5$, if I understand correctly), in particular for $n=3$.

We have $$\sum_k\frac{(-1)^kP(nk)}{k}=\sum_{k,p}\frac{(-1)^k}{kp^{nk}}=-\sum_p\ln\left(1+\frac{1}{p^n}\right)=\sum_p\ln\left(\frac{1-\frac{1}{p^{n}}}{1-\frac{1}{p^{2n}}}\right)=\ln\frac{\zeta(2n)}{\zeta(n)}. $$

This computation shows that your guess is correct whenever the numerator of $\zeta(2n)/\pi^{2n}$ is equal to $1$ (more or less for $n$ up to $5$, if I understand correctly), in particular for $n=3$.