 Solution

Enumerate the countable atomic model as $M=\{a_0,a_1,\dots \}$ and fix any $N\models T$. We recursively construct finite partial elementary maps $f_n$ from the first $n$ elements of $M$ into $N$.

Suppose $f_n(\bar{a})=\bar{b}$. Since $M$ is atomic, choose a formula $\theta (\bar{x},y)$ isolating $\mathrm{tp}(\bar{a},a_n)$. Its existential consequence $\exists y\theta (\bar{x},y)$ belongs to $\mathrm{tp}(\bar{a})$, so partial elementarity givesChoose a witness $b_n$. Because $\theta$ isolates the complete joint type, extending $f_n$ by $a_n\mapsto b_n$ remains partial elementary. The union of the recursive maps is an elementary embedding $M\to N$. Thus every countable atomic model is a prime model, proving the countable atomic model is prime result.