I think, here, I found $$ P_x(s)=\sum_{p < x} \frac{1}{p^s} =\sum_{n=1}^{\infty}\frac{ \mu (n)}{n} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \left( \left[ {\rm li}(t^{\frac zn-s}) \right]^{x}_2 -\left( t^{-s}\operatorname{li}(t^{\frac zn}) \right)_2^x \right) $$

where $\rho$ are all the zeros (trivial and non-trivial) of $\zeta$ function. See the linked question for more detail, corrections are welcome. Further we know, that $$ P(s)=\sum_{n 0}\frac {\mu(n)}n{\log\zeta(ns)} . $$

So my question is

If $\lim_{x\to \infty} P_x(s)=P(s) $ then $$ \log\zeta(ns)=\lim_{x\to \infty} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \left( \left[ {\rm li}(t^{\frac zn-s}) \right]^{x}_2 -\left( t^{-s}\operatorname{li}(t^{\frac zn}) \right)_2^x \right) ? $$