 Solution

Assume $M$ is a prime model. By the downward Lowenheim-Skolem theorem, $T$ has a countable model, and the elementary embedding of $M$ into it makes $M$ countable. If a tuple $\bar{a}\in M$ had a nonisolated type, the omitting types theorem would give a countable model of $T$ omitting that type. An elementary embedding of $M$ into this model would realize it, a contradiction. Thus $M$ is atomic.

Conversely, let $M$ be countable and atomic, enumerate it as $(a_i{)}_{i<\omega }$, and let $N\models T$. Construct an elementary embedding recursively. Suppose $a_0,\dots ,a_{n-1}$ have been mapped to $\bar{b}$. Let $\psi (\bar{x})$ isolate the type of $(a_0,\dots ,a_{n-1})$ and let $\theta (\bar{x},y)$ isolate the type of $(a_0,\dots ,a_n)$. Since the latter extends the former and is realized in $M$, completeness givesThe tuple $\bar{b}$ realizes $\psi$, so a suitable image of $a_n$ exists in $N$. The union of the finite partial elementary maps is an elementary embedding $M\to N$. Hence $M$ is prime.