Let me first lead with that the question, as stated, is nearly a duplicate of this question. However, since this question also asks for more general approaches to problems like this, I'll leave a relatively general approach below that allows one to calculate series like this without too much effort.
Approach
We utilize Riemann-Stieltjes integration. Observe that
$$
\sum_{p\text{ prime}}f(p)=\int_{2^{-}}^\infty f(x)\,d\pi(x)
$$
where $\pi(x)=\sum_{p\text{ prime}}1$ is the prime counting function. Thus, if we find a good approximation of $\pi(x)$, we can estimate the value of integrals like this asymptotically using integration by parts. For several problems, the approximation $$\pi(x)=\frac{x}{\log x}+O\left(xe^{-c\sqrt{\log x}}\right)$$ for a positive constant $c$ (better estimates are known but this is sufficient for our purposes). For $p\leq T$, we have
\begin{align*}
\int_{2^-}^T f(x)\,d\pi(x)&=\int_{2^-}^T f(x)\,d\left(\frac{x}{\log x}+O\left( xe^{-c\sqrt{\log x}}\right)\right)\\
&=\int_{2^-}^Tf(x)\,d\left(\frac{x}{\log x}\right)+\int_{2^{-}}^T f(x)\,d\left(O\left( xe^{-c\sqrt{\log x}}\right)\right)\\
&=\int_2^T f(x)\left(\frac{\log x -1}{\log^2 x}\right)\,dx+\mathcal{E}\tag{1}
\end{align*}
where $\mathcal{E}$ represents the error. The error can be estimated using integration by parts in the following way:
\begin{align*}
\int_{2^-}^T f(x)\,dO\left( xe^{-c\sqrt{\log x}}\right)&\ll xf(x)e^{-c\sqrt{\log x}}\bigg]_{x=2}^{x=T}+\int_{2}^T xe^{-c\sqrt{\log x}}\ f'(x)\,dx\\
&=O\left(\max\left\{Tf(T)e^{-c\sqrt{\log T}},1\right\}\right)+\int_2^{T/2} f'(x) e^{-c\sqrt{\log x}}\,dx \\ & \quad +\int_{T/2}^T f'(x)e^{-c\sqrt{\log x}}\,dx\\
&\ll \max\left\{f(T)e^{-c\sqrt{\log T}},1\right\}
\end{align*}
Here, I've skipped a few relatively minor details because it's dependent on the specific function $f$ that you are summing. First: the constant $c$ in the exponent changes (but remains bounded below by a positive constant). The split of the integral in the second step is not always strictly necessary, but there are instances where it is beneficial, so I wanted to write it down here. Lastly, the first integral in the second step has not been evaluated whereas the second integral can be bounded above using the lower bound $T/2$ for the exponential term and the fundamental theorem of calculus.
Applying the Approach to this Problem
In this problem, we have $f(p)=\frac{\log p}{p}$, or in the continuous analog, $f(x)=\frac{\log x}{x}$. Inserting this into the integral, we need to calculate the main term
$$
\int_2^T\frac{\log x-1}{x\log x}\,dx=\int_2^T\frac{1}{x}-\frac{1}{x\log x}\,dx
$$
Integrating, we have the main term equal to
$$
\log T-\log\log T+O(1)
$$
I'll leave verifying the error terms for you, but this shows the following asymptotic:
$$
\sum_{p\leq x\\ p\text{ prime}}\frac{\log p}{p}\sim \log x
$$
Riemann's Hypothesis
One of the reasons RH is frequently assumed in analytic number theory papers is because the estimate for primes becomes the much better
$$
\pi(x)=\int_2^x {\frac{{dt}}{{\log t}}}+O(\sqrt{x}\log x).
$$
It is easy to see that
$$
\sum\limits_p {\frac{1}{{p(p - 1)}}} = \sum\limits_{k = 2}^\infty {P(k)} = \sum\limits_{n = 1}^\infty {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^\infty {\log \zeta (nk)} } \right)} ,
$$
where $P$ is the prime zeta function, $\zeta$ is the Riemann zeta function, and $\mu$ is the Möbius function. Then
\begin{align*}
\sum\limits_{n = 1}^\infty {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^\infty {\log \zeta (nk)} } \right)} = \;& \sum\limits_{n = 1}^N {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^{\left\lceil {N/n} \right\rceil + 1} {\log \zeta (nk)} } \right)} \\ & + \sum\limits_{n = N + 1}^\infty {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^\infty {\log \zeta (nk)} } \right)} \\ &+ \sum\limits_{n = 1}^N {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = \left\lceil {N/n} \right\rceil + 2}^\infty {\log \zeta (nk)} } \right)}
\end{align*}
for any $N\geq 1$. Now
$$
\log \zeta (nk) \le 2 \cdot \frac{1}{{2^{nk} }}
$$
for any $n\ge 1$ and $k\ge 2$. Thus for any $n\ge 1$ and $M\ge 1$,
$$
\sum\limits_{k = M + 1}^\infty {\log \zeta (nk)} \le 2 \cdot \sum\limits_{k = M + 1}^\infty {\frac{1}{{2^{nk} }}} = 2^{1 - nM} .
$$
Therefore
$$
\left| {\sum\limits_{n = N + 1}^\infty {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^\infty {\log \zeta (nk)} } \right)} } \right| \le 2\sum\limits_{n = N + 1}^\infty {\frac{1}{{n \cdot 2^n }}} \le 2\int_N^{ + \infty } {\frac{{{\rm d}t}}{{t \cdot 2^t }}} \le \frac{2}{{\log 2}}\frac{1}{{N \cdot 2^N }}
$$
and
$$
\left| {\sum\limits_{n = 1}^N {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = \left\lceil {N/n} \right\rceil + 2}^\infty {\log \zeta (nk)} } \right)} } \right| \le 2\sum\limits_{n = 1}^\infty {\frac{1}{{n \cdot 2^{n\left\lceil {N/n} \right\rceil + n} }}} \le 2\frac{1}{{2^N }}\sum\limits_{n = 1}^\infty {\frac{1}{{n \cdot 2^n }}} = \frac{{2\log 2}}{{2^N }}
$$
for any $N\ge 1$. Accordingly,
$$
\sum\limits_p {\frac{1}{{p(p - 1)}}} = \sum\limits_{n = 1}^N {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^{\left\lceil {N/n} \right\rceil + 1} {\log \zeta (nk)} } \right)} + R_N ,
$$
for any $N\ge 1$, where
$$
\left| {R_N } \right| \le \frac{2}{{\log 2}}\frac{1}{{N \cdot 2^N }} + \frac{{2\log 2}}{{2^N }} < \frac{5}{{2^N }}.
$$
With $N=3400$, I got
\begin{align*}& 0.77315666904979512786436745985594239561874133608318604831100606735670\\&
9028489233397833798758823320818328937814256148184877115780660449046900\\&
2378832590763500601782383946617309470017305607667464101342674273541621\\&
9952460707589642449366317124379893255273683122617198789965614988900346\\&
8789581761777440507530380403930615725533887927380132834996433277138637\\&
7049479823191921992768751966642321281245994223008799785641115700164415\\&
4003845978165611336094017461569027505458605345995912204829815522858359\\&
0259088070578636499788833417902629524916542777136670298457774510084876\\&
5454067301120053261899310892066356658770246502596716068326040486612803\\&
6312902391262059228007144014989942678635902481893404222498689320321229\\&
1181268463226742700101487927499658707519455860084392737464646754241323\\&
4741957037164652252932590566930054867393817757225596885032596671453783\\&
9155845083333132992637494356480167180297031616517608172434964548568691\\&
0734031704796454256219160251853205082189649435393816591198262102370390\\&
29428059110079374440938\ldots
\end{align*}
for the first $1000$ digits.
Best Answer
Well, $\frac{m \log m}{2} \le p_m \le 2 m \log m$. So $\sum_{k=1}^m \frac{1}{p_k}$ satisfies
$$\ln \ln (m \log m/2) \approx \sum_{p \le \frac{m \log m}{2}} \frac{1}{p} \le \sum_{k=1}^m \frac{1}{p_k} \le \sum_{p \le 2m \log m} \frac{1}{p} \approx \ln \ln (2m \log m )$$.
However, $\ln \ln (m \log m/2)$ $\approx \ln \ln m \approx$ $\ln \ln (2m \log m )$, yielding
$$\sum_{k=1}^m \frac{1}{p_k} \approx \ln \ln m.$$
And it makes sense: From analysis (forget about primes and their distributions for a moment)
$\sum_{k=1}^N \frac{1}{k \ln k} \approx \ln \ln N$ (and $p_k \approx k \ln k$); $\sum_{k=1}^N \frac{1}{k \ln k \ln \ln k} \approx \ln \ln \ln N$ while $\sum_{k=1}^{\infty} \frac{1}{k \ln^2 k}$ is finite. In fact, so is $\sum_{k=1}^{\infty} \frac{1}{k (\ln k) (\ln \ln k)^2}$. And even, $\sum_{k=1}^{\infty} \frac{1}{k (\ln k) (\ln \ln k) (\ln \ln \ln k)^2}$ is finite too.
If this is still surprising to you, let $Y(x) = e^{e^x}$, then $dY/dx = Y \ln Y$. And let $Z = e^{e^{e^x}}$. Then $dZ/dx = Z \ln Z \ln \ln Z$.