It can be shown that
$$\sum_{n=1}^\infty\frac{P(ns)}{n}k^n=\log\left(\sum_{n=1}^\infty\frac{k^{\Omega(n)}}{n^s}\right)$$ and $$\zeta^{k}(s)=\left(\sum_{n=1}^\infty\frac{\binom{i_1+k-1}{k-1}\cdots\binom{i_r+k-1}{k-1}}{n^s}\right)$$
where $P(s)$ is the prime zeta function, $\Omega(n)$ is the total number of prime factors of $n$, and $i_r$ represents the multiplicity of the $r$th prime factor of $n$.
Now suppose for some function $h$ we had
$$\sum_{n=1}^\infty\frac{P(ns)}{n}k^n=\log\left(\sum_{n=1}^\infty\frac{k^{\Omega(n)}}{n^s}\right)=\log\zeta^{h(k)}(s)=\log\left(\sum_{n=1}^\infty\frac{\binom{i_1+h(k)-1}{h(k)-1}\cdots\binom{i_r+h(k)-1}{h(k)-1}}{n^s}\right)$$
By equating the coefficients in the dirichlet series, this problem reduces to finding for what function $h$,
$$k^{\Omega(n)}=\binom{i_1+h(k)-1}{h(k)-1}\cdots\binom{i_r+h(k)-1}{h(k)-1}$$
Of course it would be easy if we wrote the inverse of the equation, substituting $k$ with $h(k)$ and vice versa and solving; but $h(k)$ should not be in terms of $n$. So perhaps equating the dirichlet coefficients is not simplifying the problem at all. I'll just take any thoughts on if whether it's possible and what to try.