Well, one can always say that the PNT is equivalent to $$\sum_{p \leq x}\frac{1}{p} = \log \log x + M + o(\log x),\tag{$\ast$}$$ because both results are true (with better error terms). This is of course not what is meant by the Wikipedia page. Instead, the idea is that the equivalence PNT$\,\Leftrightarrow(\ast)$ can be established in a simpler way than either PNT or $(\ast)$. On the other hand, "simpler" is a subjective word, e.g. I usually find Tauberian arguments tricky.

At any rate, the first three exercises in Section 8.1.1 of Montgomery-Vaughan: Multiplicative number theory I address this question. For example, the PNT easily implies the relation $\psi(x)\sim x$ (logarithmic weights!), which then implies rather nontrivially (using Theorem 8.1 = Axer's theorem) that $$\sum_{p\leq x}\frac{\log p}{p}=\log x+C+o(1)$$ for some constant $C$. From here, $(\ast)$ follows easily by partial summation.

Well, one can always say that the PNT is equivalent to $$\sum_{p \leq x}\frac{1}{p} = \log \log x + M + o\left(\frac{1}{\log x}\right),\tag{$\ast$}$$ because both results are true (with better error terms). This is of course not what is meant by the Wikipedia page. Instead, the idea is that the equivalence PNT$\,\Leftrightarrow(\ast)$ can be established in a simpler way than either PNT or $(\ast)$. On the other hand, "simpler" is a subjective word, e.g. I usually find Tauberian arguments tricky.

At any rate, the first three exercises in Section 8.1.1 of Montgomery-Vaughan: Multiplicative number theory I address this question. For example, the PNT easily implies the relation $\psi(x)\sim x$ (logarithmic weights!), which then implies rather nontrivially (using Theorem 8.1 = Axer's theorem) that $$\sum_{p\leq x}\frac{\log p}{p}=\log x+C+o(1)$$ for some constant $C$. From here, $(\ast)$ follows easily by partial summation.