Just to elaborate a bit in explicitly differential-geometric terms on MTS's answer, which refers to certain results of Plymen's originally restated in terms of Morita equivalence (via the dictionary given by the Serre–Swan theorem), let $M$ be a compact orientable Riemannian manifold, and let $\operatorname{\mathbb{C}l}^{(+)}(M)$ be the finite rank Azumaya bundle given by the complexification of the Clifford bundle $\operatorname{Cl}(M)$ if $\dim M$ is even, and by the complexification of the even subbundle of the Clifford bundle if $\dim M$ is odd. Then $M$ is spin$^\mathbb{C}$ if and only if there exists an irreducible $\operatorname{\mathbb{C}l}^{(+)}(M)$-module, that is, a Hermitian vector bundle $\mathcal{S} \to M$ (i.e., a spinor bundle) such that $\operatorname{\mathbb{C}l}^{(+)}(M) \cong \operatorname{End}(\mathcal{S})$.
Now, if what you care about are specifically spin manifolds, one can endow $\operatorname{\mathbb{C}l}^{(+)}(M)$ with a canonical $\mathbb{C}$-linear anti-involution, and hence equip the dual bundle $\mathcal{E}^*$ of a $\operatorname{\mathbb{C}l}^{(+)}(M)$-module with the structure of a $\operatorname{\mathbb{C}l}^{(+)}(M)$-module. It is then , that $M$ is actually spin if and only if there exists an irreducible $\operatorname{\mathbb{C}l}^{(+)}(M)$-module $\mathcal{S} \to M$ such that $\mathcal{S} \cong \mathcal{S}^\ast$ not only as Hermitian vector bundles, but also as $\operatorname{\mathbb{C}l}^{(+)}(M)$-modules—this is, then, the spinor bundle for its corresponding spin structure. Indeed, by the anti-unitary isomorphism $\mathcal{S} \cong \mathcal{S}^\ast$ of Hermitian vector bundles defined by the Hermitian metric, together with a little bit of care, one can recognise such a unitary isomorphism of $\operatorname{\mathbb{C}l}^{(+)}(M)$-modules as nothing else than the charge conjugation operator on spinors in mild disguise.