 Solution

If a model $B$ of $T$ contains $A$ as a substructure, every existential sentence true in $A$ remains true in $B$: use the same witness tuple and preservation of quantifier-free truth. Hence $B$ satisfies the required joint theory.

For the existential-theory embedding criterion, conversely suppose that joint theory has a model. Expand the language by a constant $c_a$ for each $a\in A$, and add the atomic and negated atomic diagram $D(A)$ to $T$. Any finite collection of its literals is a quantifier-free conjunction $\theta (\bar{c})$ true of the corresponding tuple in $A$. Replacing those named constants by variables and existentially closing gives a sentence in ${\mathrm{Th}}_{\exists }(A)$. A model of the assumed joint theory can supply its witnesses, and hence an expansion satisfying those finitely many diagram requirements and any finite subset of $T$.

Compactness therefore gives a model of $T+D(A)$. Its named Elements define an embedding of $A$ into its $L$-reduct: inequalities ensure injectivity, and all function and relation literals ensure preservation and reflection. Replace the ambient model by an isomorphic copy identifying this image with $A$ itself. Thus a model of $T$ containing $A$ exists. The diagram step explains how parameter-free existential information nevertheless yields an embedding of the whole structure.