From $\varphi(n) = \sum_{d | n} \mu(d) \frac{n }{d}$ $$\sum_{n=1}^\infty \varphi(n) n^{-s} =\frac{\zeta(s-1)}{\zeta(s)}$$ Since $\frac{\zeta(s-1)}{\zeta(s)}$ has a dominating pole of order $1$ at $s=2$ and $\varphi(n) \ge 0$ we obtain $$\sum_{n < x} \varphi(n) \sim \frac{x^2}{\zeta(2)}$$ The PNT means $p_n \sim n \ln n$ so your asymptotic is not correct.