Does sum of the reciprocals of all the composite numbers converge?

Since $\sum_{n=1}^\infty \frac{1}{n}$ diverges, so does $\sum_{n=2}^\infty\frac{1}{2n}$. All the terms in this series are included in your series, so your series also diverges.

Simple answer : For all prime $p$ (except for 2 and 3), $p-1$ is composite, so $\sum_{p\in\mathcal P-\{2,3\}}\frac 1 p < \sum_{p\in\mathcal P-\{2,3\}}\frac 1 {p-1} < \sum_{n\notin\mathcal P}\frac 1 n$

This sum does not converge. Consider for example the sum of the reciprocals of all the even numbers greater than 2: $1/4 + 1/6 + 1/8 + 1/10 + \ldots$. This diverges, the proof being analogous to the proof that the harmonic series diverges. Thus your series has arbitrarily large partial sums and thus diverges.