The constant $$\alpha = \sum_{p} \frac{\log \left({p}\right)}{p \left({p-1}\right)}$$ comes from the calculation $$\sum_{p=2}^{x} \frac{\log \left({p}\right)}{p-1} = \sum_{p=2}^{x} \frac{\log \left({p}\right)}{p} + \alpha$$
Where $p$ is a prime. Using the primes up to ${10}^{9}$ results in $\alpha = 0.75536660\cdots$. Does this constant have a name and is it listed with additional decimal digits of value.
The only reference to date where the constant is computed is Explicit estimates for the Artin L-functions: Duke's short-sum theorem and Dedekind zeta residues, by Stephan Ramon Garcia and Ethan Simpson Lee at arXiv:2101.11853v5 section 2.5
Using the results below where $\alpha$ is summed over Mobius and zeta function results in a faster convergence where we can write out to 30 decimal digits $$\alpha = 0.755366610831688021159316685989\, \cdots$$
(Incomplete answer)
Define the Prime Zeta Function $P(s)=\sum_{p}\frac{1}{p^s}$. This converges uniformly for $\Re(s)>1$. Taking the derivative, $$P'(s)=-\sum_p\frac{\log p}{p^s}.$$ Then $$\begin{align} \alpha&=\sum_p\frac{\log p}{p(p-1)}\\ &=\sum_p\frac{\log p}{p}\frac{1/p}{1-1/p}\\ &=\sum_p\frac{\log p}{p}\sum_{n\ge1}\frac{1}{p^n}\\ &=\sum_{n\ge1}\sum_p\frac{\log p}{p^{n+1}}\\ &=-\sum_{n\ge1}P'(n+1)\\ &=-\sum_{n\ge2}P'(n). \end{align}$$ Now, recall the Euler product for $\zeta(s)$: $$\zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}.$$ Taking the logarithm, $$\begin{align} \log\zeta(s)&=-\sum_p\log(1-p^{-s})\\ &=\sum_p\sum_{n\ge1}\frac{p^{-ns}}{n}\\ &=\sum_{n\ge1}\frac{1}{n}\sum_p\frac{1}{p^{ns}}\\ &=\sum_{n\ge1}\frac{P(ns)}{n}, \end{align}$$ This can be re-written as $$-\sum_{n\ge2}\frac{P(ns)}{n}=P(s)-\log\zeta(s).$$ Upon differentiating, $$-\sum_{n\ge2}P'(ns)=P'(s)-\frac{\zeta'(s)}{\zeta(s)},$$ thus $$\alpha=-\sum_{n\ge2}P'(n)=\lim_{s\to 1^{+}}\left(P'(s)-\frac{\zeta'(s)}{\zeta(s)}\right).$$ Now, via Mobius inversion, $$\log\zeta(s)=\sum_{n\ge1}\frac{P(ns)}{n}\iff P(s)=\sum_{n\ge1}\frac{\mu(n)}{n}\log\zeta(ns).$$ Differentiating $P(s)$, $$P'(s)=\sum_{n\ge1}\mu(n)\frac{\zeta'(ns)}{\zeta(ns)}=\frac{\zeta'(s)}{\zeta(s)}+\sum_{n\ge2}\mu(n)\frac{\zeta'(ns)}{\zeta(ns)}.$$ So, $$\alpha=\lim_{s\to 1^{+}}\left(P'(s)-\frac{\zeta'(s)}{\zeta(s)}\right)=\lim_{s\to 1^{+}}\sum_{n\ge2}\mu(n)\frac{\zeta'(ns)}{\zeta(ns)}=\sum_{n\ge2}\mu(n)\frac{\zeta'(n)}{\zeta(n)}.$$ As far as I know there is no known closed form for this.