UPD. GH's approach is, of course, better, the reason is that it uses a stronger asymptotical estimate: asymptotics $\sum_{p\leqslant x} \log p/p=\log x+O(1)$ implies $\sum_{p\leqslant x} 1/p=\log\log x+O(1)$, but not viceversa.

We use $$F(x):=\sum_{p\leqslant x} \frac1{p}=\log\log x+O(1),$$
see here about more precise statement.
Then apply Abel transform, for $h(p)=\log\log p$:
$$
\sum_{p\leqslant x}\frac{h(p)}{p}=\sum_{n\leqslant x} h(n)(F(n)-F(n-1))=h([x])F([x])-\sum_{n\leqslant x-1} F(n)(h(n+1)-h(n))
$$
First guy $h([x])F([x])$ equals $h^2(x)+O(h(x))$. As for the subtracted term, at first we replace each $F(n)$ to $h(n)+O(1)$, then sum of errors is
$$
O\left(\sum_{n\leqslant x-1} (h(n+1)-h(n))\right)=O(h(x)).
$$
Next, for $h(n)(h(n+1)-h(n))$, it equals
$$
h(n)(h(n+1)-h(n))=\frac{(h(n+1))^2-(h(n))^2}2-
\frac{(h(n+1)-h(n))^2}{2}.
$$
First terms cancel telescopically and give $\frac{h^2(x)}{2}+O(1)$, the second terms are really small and give just $O(1)$.

To summarize, the answer is $\frac12 h^2(x)+O(h(x))$.