Let $c : \mathbb{C}\mathrm{l}_n \to \operatorname{End}(S_n)$ be your irrep, so that the “dual” irrep $c^\ast : \mathbb{C}\mathrm{l}_n \to \operatorname{End}(S_n^\ast)$ is defined by
$$
\forall x \in \mathbb{C}\mathrm{l}_n, \quad c^\ast(x) := c(x^!)^t.
$$
Thus, if $\pi := c\vert_{\operatorname{Spin}^\mathbb{C}(n)} : \operatorname{Spin}^\mathbb{C}(n) \to U(S_n)$, then $\pi^\ast : \operatorname{Spin}^\mathbb{C}(n) \to U(S_n^\ast)$ is given by
$$
\forall \lambda \in \mathbb{T}, \; \forall x \in \operatorname{Spin}(n), \quad \pi^\ast(\lambda x) = c((\lambda x)^{-1})^t = c^\ast(\lambda^{-1}x),
$$
so that the induced representation $$\operatorname{Hom}(\pi,\pi^\ast) : \operatorname{Spin}^\mathbb{C}(n) \to U(\operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast)) = \mathbb{T}\operatorname{id}_{\operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast)}$$ is given by
$$
\forall \lambda \in \mathbb{T}, \; \forall x \in \operatorname{Spin}(n), \; \forall T \in \operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast), \\ \operatorname{Hom}(\pi,\pi^\ast)(\lambda x)T := \pi^\ast(\lambda x) \circ T \circ \pi(\lambda x)^{-1} = c^\ast(\lambda^{-1} x) \circ T \circ c(\lambda^{-1}x^!) = c^\ast\left(\nu((\lambda x)^{-1})\right) \circ T.
$$
Since
$$
S \cong \operatorname{Spin}(E) \times_\pi S_n, \quad S^\ast \cong \operatorname{Spin}(E) \times_{\pi^\ast} S_n^\ast,
$$
you can now use local trivialisations of $\operatorname{Spin}(E)$ to show that
$$
\operatorname{Hom}_{\mathbb{C}\mathrm{l}(E)}(S,S^\ast) \cong \operatorname{Spin}(E) \times_{\nu^{-1}} \mathbb{C} \cong L_E^\ast,
$$
and hence, by the natural isomorphism $S_n^\ast \cong S_n \otimes \operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast)$, that
$$
S^\ast \cong S \otimes \operatorname{Hom}_{\mathbb{C}\mathrm{l}(E)}(S,S^\ast) \cong S \otimes L_E^\ast.
$$