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Use the countable-language hypothesis of the omitting types theorem, which is the usual convention for this prime-model result. Let $\mathcal{M}$ be a prime model of $T$ and $a\in M$.

First, $T$ is complete up to logical consequence: every model $\mathcal{N}\models T$ receives an elementary embedding of $\mathcal{M}$, so it satisfies exactly the same sentences as $\mathcal{M}$. We may therefore replace $T$ by $\mathrm{Th}(\mathcal{M})$ without changing its model class or the prime model property.

Suppose $p={\mathrm{tp}}^{\mathcal{M}}(a/\varnothing )$ were not isolated. The omitting types theorem, applied to this single complete type, would produce $\mathcal{N}\models T$ omitting $p$. By primality choose an elementary embedding $j:\mathcal{M}\to \mathcal{N}$. For every $\phi (x)\in p$,Then $j(a)$ realizes $p$, contradicting omission. ConsequentlyThe same proof applies to every finite tuple, giving an atomic model. Countability of the language is a hypothesis of this argument, not a restriction on parameter sets in the separate parts (a) and (d). The PDF does not state that standing hypothesis explicitly; it must be included when invoking this version of the theorem.

A prime model of a complete theory $T$ is a model $M\models T$ for which there is an elementary embedding $M\to N$ into every model $N\models T$.

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.

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.