The prime zeta function has the following expansion near its singularity at 1:
$$P(1+\varepsilon) = -\ln \varepsilon + C + O(\varepsilon)$$
It also has a singularity at the reciprocal of every square-free positive integer. What is its expansion near each of these singularities? Is it
$$P(1/n+\varepsilon) = -\frac{\mu(n)}{n} \ln \varepsilon + C + O(\varepsilon) $$
where $\mu$ is the Möbius function?
There are two main ingredients here. The first is $\zeta$ near its singularity $s=1$:
$$ \zeta(1+\varepsilon)=\frac{1}{\varepsilon}+\gamma+\mathcal{O}(\varepsilon) $$
(where $\gamma$ is the Euler-Mascheroni constant). The other is the Mobius inversion formula:
$$ P(s)=\sum_{k=1}^\infty \frac{\mu(k)}{k}\ln \zeta(ks). $$
This is mentioned on the MathWorld page you linked for the prime zeta function $P(s)$. Notice the only nonzero terms occur when $k$ is squarefree. Setting $s=1/n+\epsilon$ yields
$$ P(1/n+\epsilon)=\frac{\mu(n)}{n}\ln\zeta(1+n\varepsilon)+\sum_{\substack{k=1\\ k\ne n}}^{\infty} \frac{\mu(k)}{k}\ln\zeta(k(1/n+\varepsilon)). $$
Letting $\varepsilon\to0$ in this case, and using the $\mathcal{O}$ expansion of $\zeta$ and $\ln$ yields
$$ \begin{array}{ll} \ln\zeta(1+n\varepsilon) & =\ln\left(\frac{1}{n\varepsilon}+\gamma+\mathcal{O}(\varepsilon)\right) \\ & = -\ln(n\epsilon)+\ln(1+n\gamma\epsilon+\mathcal{O}(\varepsilon^2)) \\ & = -\ln(n\varepsilon)+\mathcal{O}(\varepsilon). \end{array} $$
Therefore,
$$ P(1/n+\varepsilon)=-\frac{\mu(n)}{n}\ln(\varepsilon)+\left[-\frac{\mu(n)}{n}\ln(n)+\sum_{\substack{k=1\\k\ne n}}^\infty \frac{\mu(k)}{k}\ln\zeta(k/n)\right]+\mathcal{O}(\varepsilon). $$