Definition A theory $T$ is a set of sentences. A structure $\mathcal{A}$ is a model of $T$ if $\mathcal{A}\vDash T$.
To better understand the situation, let us recall the classical Galois connection machinery used in these situations. Fix a language $\mathcal{L}$; with a set of formulae $T$ one can associate the class of $\mathcal{L}$-structures $\operatorname{Mod}(T)=\{\mathcal{A}: \mathcal{A}\vDash T\}$ and conversely with a class of $\mathcal{L}$-structures $\mathcal{C}$ one can associate a set of formulae $\operatorname{Th}(\mathcal{C})=\{B: \textrm{for all }\mathcal{A}\in\mathcal{C},\mathcal{A}\vDash B\}$. It is quite immediate to show that we have $$T\subseteq \operatorname{Th}(\mathcal{C})\Leftrightarrow\mathcal{C}\subseteq\operatorname{Mod}(T)$$ for all $T,\mathcal{C}$.
Definition A theory $T$ is said to be deductively closed if $T=\operatorname{Th}(\operatorname{Mod}(T))$.
What I can't understand is what follows: a theory $T$ is deductively closed if and only if it is of the kind $\operatorname{Th}(\mathcal{C})$ for some $\mathcal{C}$. One implication is trivial (take $\mathcal{C}:=\operatorname{Mod}(T))$. For the other one I'd need some help.
Note that for any $T$, $T \subseteq \mbox{Th}(\mbox{Mod}(T))$ (which just says all the sentences in $T$ in hold in a model of $T$). Now assume $T = \mbox{Th}({\cal C})$. Then ${\cal C} \subseteq \mbox{Mod}(T)$ (which just says that every member of $\cal C$ is a model of $T$). So as the Galois connection reverses inclusions, we have $\mbox{Th}({\cal C}) \supseteq \mbox{Th}(\mbox{Mod}(T))$, so that:
$$T = \mbox{Th}({\cal C}) \supseteq \mbox{Th}(\mbox{Mod}(T)) \supseteq T.$$