There is a general theorem, which can be proved by using Schur's complement formula: let a matrix $A\in M_{pq}(k)$ be written blockwise, with blocks $A_{ij}\in M_q(k)$ for $1\le i,j\le p$. Assume that the blocks $A_{ij}$ commutte pairwise, so that the determinantal expression $$C=\sum_{\sigma\in\frak_p}\epsilon(\sigma)\prod_{i=1}^pA_{i\sigma(i)}$$ makes sense. Then $$\det A=\det C.$$
In your case, $C=(\det B) I_d$ and one obtains readily $\det A=(\det B)^d$.