There is already a very good answer, so this answer is just meant as a supplement.
The theory $\operatorname{Th}(\mathcal{M})$ of any structure $\mathcal{M}$ is a complete theory. So any structure $\mathcal{N}$ will be a model of that theory precisely when it is elementarily equivalent to $\mathcal{M}$. For example, if $\mathcal{M} = (\mathbb{R}, <)$, the reals with just the order symbol, then $\operatorname{Th}(\mathcal{M})$ will be the theory of dense linear orders without endpoints (DLO). Then $(\mathbb{Q}, <)$ is a model of this theory.
If we consider the complete diagram $\operatorname{Diag}(\mathcal{M})$ of $\mathcal{M}$, we get something much stronger. The point of this theory is that now any model $\mathcal{N} \models \operatorname{Diag}(\mathcal{M})$ will not only be elementarily equivalent to $\mathcal{M}$, but it will also be an elementary extension of $\mathcal{M}$. Since $\operatorname{Diag}(\mathcal{M})$ contains a constant $c_a$ for every $a \in |\mathcal{M}|$, we can define a function $f: \mathcal{M} \to \mathcal{N}$ by $f(a) = c_a^{\mathcal{N}}$. Here $c_a^{\mathcal{N}}$ is the interpretation of $c_a$ in $\mathcal{N}$. Then
$$
\mathcal{M} \models \varphi(a_1, \ldots, a_n) \quad \Longleftrightarrow \quad
\varphi(c_{a_1}, \ldots, c_{a_n}) \in \operatorname{Diag}(\mathcal{M}) \quad \Longleftrightarrow \quad
\mathcal{N} \models \varphi(c_{a_1}^{\mathcal{N}}, \ldots, c_{a_n}^{\mathcal{N}}),
$$
and this last expression is just $\mathcal{N} \models \varphi(f(a_1), \ldots, f(a_n))$. So $f$ is an elementary embedding.
Back to our example of $\mathcal{M} = (\mathbb{R}, <)$. Any model $\mathcal{N} \models \operatorname{Diag}(\mathcal{M})$ must contain a copy of $\mathbb{R}$. So $(\mathbb{Q}, <)$ can no longer be a model of the complete diagram of $\mathcal{M}$, even though it was a model of the theory of $\mathcal{M}$.