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Per JDH's suggestion, I'll turn my earlier comment into an answer.

Assuming $T$ to be model-complete, then whenever $M$, $N$ and $A$ are all models of $T$, it would certainly follow from $A \subseteq M$ and $A \subseteq N$ that $M \models \phi(a)$ iff $N \models \phi(a)$ for any $a$ from $A$ (as whenever one model of $T$ is a substructure of another, it is in fact an elementary substructure). But in Marker's condition, $A$ can be any $L$-structure and is not required to be a model of $T$.

Any theory that has elimination of quantifiers is model-complete, but the converse is not true. Note that while Marker's Theorem 3.1.4 is stated for a theory in a language with at least one constant symbol, he notes afterward that the proof can be adapted to cover the case in which $L$ has no constant symbols; so if model-completeness were to have sufficed here, it would've implied that the false converse were true.Instead

Incidentally, one very interesting theory which is model-complete yet does not admit elimination of quantifiers is the theorem's model-theoretic characterization theory of EQ highlights the real field with exponentiation. This theory isn't known to be decidable, but MacIntyre and Wilkie showed that its greater strengthdecidability is implied by the real version of Schanuel's conjecture. (This nicely succinct postscript file of Kuhlmann's contains handy references.)

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Per JDH's suggestion, I'll turn my earlier comment into an answer.

Assuming $T$ to be model-complete, then whenever $M$, $N$ and $A$ are all models of $T$, it would certainly follow from $A \subseteq M$ and $A \subseteq N$ that $M \models \phi(a)$ iff $N \models \phi(a)$ for any $a$ from $A$ (as whenever one model of $T$ is a substructure of another, it is in fact an elementary substructure). But in Marker's condition, $A$ can be any $L$-structure and is not required to be a model of $T$.

Any theory that has elimination of quantifiers is model-complete, but the converse is not true. Note that while Marker's Theorem 3.1.4 is stated for a theory in a language with at least one constant symbol, he notes afterward that the proof can be adapted to cover the case in which $L$ has no constant symbols; so if model-completeness were to have sufficed here, it would've implied that the false converse were true. Instead, the theorem's model-theoretic characterization of EQ highlights its greater strength.