I assume that what the OP wanted to say was, given a correlation matrix $B$, find a correlation matrix $A$ that maximizes $\det(A+B)$. Let me cite here a more general theorem from which the conclusion follows trivially.
Theorem Let $A$ and $B$ be Hermitian matrices with eigenvalues $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$, respectively. Then,
\begin{equation*} \det(A+B) \le \max_{\sigma \in \mathfrak{S}_n}\prod_{i=1}^n (a_i + b_{\sigma(i)}). \end{equation*}
Since correlation matrices are Hermitian positive semidefinite, a specialization of this theorem shows us that \begin{equation*} \det(A+B) \le \prod_{i=1}^n (\lambda_i(A) + \lambda_{n-i+1}(B)), \end{equation*} where $\lambda_i(\cdot)$ is the $i$-largest eigenvalue. Thus, in particular, for a fixed $B$, we just need to pick a suitable $A$ that has the same eigenvectors as $B$ but has the largest possible eigenvalues---seems like $A=I_n$ does the job!