The problem is indeed convex if \mu_1 \geq \mu_2 \ldots \geq \mu_K \geq 0.

First, the expression

\log \frac{ \det (\mathbf{Q}+ \sum_{i=1}^k d_i \mathbf{h}_i \mathbf{h}_i ^ H )} { \det (\mathbf{Q}+ \sum_{i=1}^{k-1} d_i \mathbf{h}_i \mathbf{h}_i ^ H )} = \log \det \left( \mathbf{I} + \left(\mathbf{Q} + \sum_{i=1}^{k-1} d_i \mathbf{h}_i \mathbf{h}_i ^ H \right)^{-1} d_k \mathbf{h}_k \mathbf{h}_k ^ H \right)

is convex to \mathbf{Q}, so the function is convex to \mathbf{Q}.

To show that the function is concave to \mathbf{d}, we can use the following equation

\log \frac{ \det (\mathbf{Q}+ \sum_{i=1}^k d_i \mathbf{h}_i \mathbf{h}_i ^ H )} { \det (\mathbf{Q}+ \sum_{i=1}^{k-1} d_i \mathbf{h}_i \mathbf{h}_i ^ H )} = \log \det (\mathbf{Q}+ \sum_{i=1}^k d_i \mathbf{h}_i \mathbf{h}_i ^ H ) - \log \det (\mathbf{Q}+ \sum_{i=1}^{k-1}d_i \mathbf{h}_i \mathbf{h}_i ^ H ),

and rewrite the function to be

\sum_{k=1} ^ {K-1} (\mu_k - \mu_{k+1}) \log \det (\mathbf{Q}+ \sum_{i=1}^k d_i \mathbf{h}_i \mathbf{h}_i ^ H) + \mu_K \log \det ( \mathbf{Q}+ \sum_{i=1}^K d_i \mathbf{h}_i \mathbf{h}_i ^ H ) - \mu_1 \log \det (\mathbf{Q}).