This was an exercise using the first Chebyshev function, $\vartheta(x)= \sum_{p \leq x} \log p.$ The question is simply how to prove (2) below, the rest is my two thoughts on how to proceed. [Edit: trimmed after answer.]
$p_k$ is a number with k prime factors including repetitions. It is not the kth prime. $\pi_k(x)$ is the number of $p_k \leq x.$ "log" is the natural log but easier to read than "$\ln$" I think.
We know that $\sum_{p \leq x} \log p \sim x $ [Hardy and Wright p. 451]. We also know that $\sum_{n\leq x} \log n \approx \int_2^x \log t\hspace{2mm}dt \approx x \log x - x.$
So (omitting the indices),
$\frac{\sum \log p}{\sum \log n} \approx \frac{\sum \log p}{x\log x - x}\sim \frac{x}{x\log x - x }\sim \frac{1}{\log x } \sim \frac{\pi(x)}{x}.$
So we could ask whether
$(1).....\hspace{5mm}\frac{\sum \log p_k}{\sum \log n}\hspace{2mm} ?\sim \frac{\pi_k(x)}{ x}.$
For example: for x = 10000, $\pi_5 = 963,$ so $\frac{\pi_5(x) }{x} = 0.0963$ and $\frac{\sum \log p_5}{\sum \log 10^4} \approx 0.0974$
So from (1) via the prior line we would get $(2)...\hspace{4mm}\frac{\pi_k(x)}{\sum \log p_k} ? \sim \frac{1}{\log x}.$
Since $ \sum_k \pi_k(x) = x,$ we would have
$(3)...\hspace{4mm}\sum_k \sum_{p_{k}(m) \leq x} \log p_{k}(m) \sim x \log x. $
In case it's not clear: for each k we sum over $m = 1,2,3,... $ with $ p_k(m) \leq x$ and then we sum over k such that $p_k \leq x$ for some m.
So for $\pi_2(x) \approx \frac{\sum \log p_2}{\log x}$ I thought to compare it to the generalized PNT... [Edit: example omitted]
Or from (2) we could get that $\frac{\pi_k(x) = m}{\sum_{n=1}^m \log p_k(n)}\sim \frac{1}{\log x},$ so
$ \frac{(1/m) m }{(1/m) \log \prod p_k(m)}\sim \frac{1}{\log x} $ and
$ \sqrt[m]{p_k(1)p_k(2)...p_k(m)} \sim x \hspace{4mm}...?$
Suggestions/corrections welcome.