= Solution
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 $\operatorname{Th}(\mathcal M)$ without changing its model class or the <prime model> property.
Suppose $p=\operatorname{tp}^{\mathcal M}(a/\varnothing)$ 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 $\varphi(x)\in p$,
$$
\mathcal M\models\varphi(a)\quad\Longrightarrow\quad\mathcal N\models\varphi(j(a)).
$$
Then $j(a)$ realizes $p$, contradicting omission. Consequently
$$
\boxed{\operatorname{tp}^{\mathcal M}(a/\varnothing)\text{ is isolated for every }a\in M.}
$$
The 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.
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