[proofplan] We prove the two directions directly from the definition of model completeness. If $T$ is model complete, then every model of $T\cup\operatorname{Diag}(M)$ contains an elementary copy of $M$, so all such expansions agree on every sentence with constants from $M$. Conversely, if every diagram theory $T\cup\operatorname{Diag}(M)$ is complete, then an embedding $f:M\to N$ between models of $T$ makes $N$ an $L(M)$-model of the same complete diagram theory as $M$, so truth of all formulas with parameters from $M$ is preserved by $f$. [/proofplan] [step:Fix the diagram notation] For an $L$-structure $M$, the language $L(M)$ adds a constant symbol $c_m$ for each element $m\in M$. The quantifier-free diagram $\operatorname{Diag}(M)$ is the set of all quantifier-free $L(M)$-sentences true in $M$ when $c_m$ is interpreted as $m$. If $N^*$ is an $L(M)$-structure satisfying $\operatorname{Diag}(M)$, then the assignment \begin{align*} m\longmapsto c_m^{N^*} \end{align*} is an $L$-embedding from $M$ into the $L$-reduct of $N^*$. Indeed, the quantifier-free diagram records all atomic and negated atomic facts about named elements of $M$, so it records equality, inequality, relations, and the values of functions on tuples from $M$. [guided] The language $L(M)$ is the language $L$ together with a name for each element of $M$. The sentence $c_m\ne c_{m'}$ belongs to $\operatorname{Diag}(M)$ whenever $m\ne m'$, and the diagram also contains the atomic facts describing all functions and relations on named tuples. Therefore any model of $\operatorname{Diag}(M)$ contains a faithful copy of $M$: send \begin{align*} m\in M \end{align*} to the interpretation of the constant $c_m$. Because the diagram preserves all quantifier-free facts, this map is an $L$-embedding. [/guided] [/step] [step:Model completeness implies completeness of each diagram theory] Assume $T$ is model complete, and fix $M\models T$. The theory \begin{align*} T\cup \operatorname{Diag}(M) \end{align*} has a model, namely the natural $L(M)$-expansion of $M$. To prove it is complete, let $N_1^*$ and $N_2^*$ be two $L(M)$-models of this theory. Let $N_1$ and $N_2$ be their $L$-reducts. By the previous step, the interpretations of the constants define $L$-embeddings \begin{align*} e_i:M\to N_i, \qquad e_i(m)=c_m^{N_i^*}, \qquad i\in\{1,2\}. \end{align*} Since $T$ is model complete and $M,N_i\models T$, each $e_i$ is elementary. Let $\sigma$ be an arbitrary $L(M)$-sentence. Only finitely many constants occur in $\sigma$, say \begin{align*} c_{m_1},\dots,c_{m_r}. \end{align*} Let $\widehat{\sigma}(z_1,\dots,z_r)$ be the corresponding $L$-formula obtained by replacing $c_{m_j}$ by the variable $z_j$. Then, for $i\in\{1,2\}$, \begin{align*} N_i^*\models \sigma \quad\Longleftrightarrow\quad N_i\models \widehat{\sigma}(e_i(m_1),\dots,e_i(m_r)) \quad\Longleftrightarrow\quad M\models \widehat{\sigma}(m_1,\dots,m_r). \end{align*} Thus $N_1^*\models\sigma$ if and only if $N_2^*\models\sigma$. Since $\sigma$ was arbitrary, all models of $T\cup\operatorname{Diag}(M)$ satisfy the same $L(M)$-sentences, so the theory is complete. [guided] Now suppose $T$ is model complete. Fix a model $M\models T$, and take two models \begin{align*} N_1^*,N_2^*\models T\cup\operatorname{Diag}(M). \end{align*} The constants give embeddings \begin{align*} e_i:M\to N_i \end{align*} into the $L$-reducts $N_i$. Because $T$ is model complete, these embeddings are elementary. To compare the two $L(M)$-models, take any $L(M)$-sentence $\sigma$. It uses only finitely many constants \begin{align*} c_{m_1},\dots,c_{m_r}. \end{align*} After replacing these constants by variables, $\sigma$ becomes an $L$-formula \begin{align*} \widehat{\sigma}(z_1,\dots,z_r). \end{align*} Elementarity of $e_i$ gives \begin{align*} N_i^*\models \sigma \quad\Longleftrightarrow\quad M\models \widehat{\sigma}(m_1,\dots,m_r). \end{align*} The right-hand side is independent of $i$, so $N_1^*$ and $N_2^*$ agree on $\sigma$. Hence the diagram theory is complete. [/guided] [/step] [step:Completeness of diagram theories implies model completeness] Conversely, assume that for every $M\models T$ the $L(M)$-theory \begin{align*} T\cup\operatorname{Diag}(M) \end{align*} is complete. Let $f:M\to N$ be an $L$-embedding between models of $T$. Expand $M$ to an $L(M)$-structure $M^*$ by interpreting $c_m$ as $m$, and expand $N$ to an $L(M)$-structure $N_f^*$ by interpreting \begin{align*} c_m^{N_f^*}:=f(m). \end{align*} Because $f$ is an embedding, $N_f^*$ satisfies $\operatorname{Diag}(M)$; and since $N\models T$, it satisfies $T$. Thus \begin{align*} M^*,N_f^*\models T\cup\operatorname{Diag}(M). \end{align*} By completeness of this diagram theory, $M^*$ and $N_f^*$ satisfy the same $L(M)$-sentences. Let $\varphi(x_1,\dots,x_r)$ be any $L$-formula and let $m_1,\dots,m_r\in M$. Applying the preceding sentence agreement to \begin{align*} \varphi(c_{m_1},\dots,c_{m_r}) \end{align*} gives \begin{align*} M\models \varphi(m_1,\dots,m_r) \quad\Longleftrightarrow\quad N\models \varphi(f(m_1),\dots,f(m_r)). \end{align*} Therefore $f$ is elementary. Since every embedding between models of $T$ is elementary, $T$ is model complete. [guided] For the converse, suppose every theory $T\cup\operatorname{Diag}(M)$ is complete. Let \begin{align*} f:M\to N \end{align*} be an embedding between models of $T$. Use $f$ to interpret each new constant $c_m$ in $N$ by \begin{align*} c_m^{N_f^*}=f(m). \end{align*} Because $f$ preserves all quantifier-free $L$-facts, this expansion $N_f^*$ satisfies $\operatorname{Diag}(M)$. The natural expansion $M^*$ also satisfies $\operatorname{Diag}(M)$. Thus both \begin{align*} M^* \qquad\text{and}\qquad N_f^* \end{align*} are models of the complete theory $T\cup\operatorname{Diag}(M)$. They agree on every $L(M)$-sentence. In particular, for each $L$-formula $\varphi(x_1,\dots,x_r)$ and each tuple $m_1,\dots,m_r\in M$, \begin{align*} M\models \varphi(m_1,\dots,m_r) \quad\Longleftrightarrow\quad N\models \varphi(f(m_1),\dots,f(m_r)). \end{align*} This is exactly the statement that $f$ is elementary. Hence every embedding between models of $T$ is elementary, so $T$ is model complete. [/guided] [/step]