We want to estimate $\displaystyle \sum_{p|n} \frac{\log p}{p}$. It's well known that $\displaystyle \sum_{p \le n} \frac{\log p}{ p} = \log n + O(1)$.
Of course the former series should be smaller than second one, but what's the behavior? I think it should be $\log \log n$ (like instead of $n$ we should consider $\log n$, because of number of elements in this sum). Also I thought about Moebius function, but it doesn't work here. Any hints?
Let $n \in \mathbb{N}^{*}$: $$\displaystyle \frac{\log n}{n} \leq \sum_{\substack{p|n \\ \text{p prime}}} \frac{\log p}{p}$$ And we have equality iff $n$ is prime.
Let $n = \displaystyle{\small \prod_{\substack{p \leq m \\ \text{p premier}}} {\normalsize p}}$. Then : $$ \begin{array}{rcl} \displaystyle\sum_{\substack{p|n \\ \text{p prime}}} \frac{\log p}{p} & = & \displaystyle\sum_{\substack{p \leq m \\ \text{p prime}}} \frac{\log p}{p} \\ \\ & = & \log m + O(1) \\ \\ & = & \log\Big((1+o(1))\log(n)\Big)\,\, , \quad (\text{prime number theorem})\end{array}$$
Then we have for all $n \in \mathbb{N}^{*}$: $$\displaystyle \frac{\log n}{n} \leq \sum_{\substack{p|n \\ \text{p prime}}} \frac{\log p}{p} \leq \log\Big((1+o(1))\log n\Big)$$
You can see that $\displaystyle\sum_{\substack{p|n \\ \text{p prime}}} \frac{\log p}{p}$ haven't an equivalent whene $n \to +\infty$.