Can the following quantity be reduced to further
$[A \otimes B + (\mathbb{1} - A) \otimes D]^N$, for positive interger $N$? Here, $\otimes$ denotes the Kronecker (tensor) product and the matrices $A$ and $B$ may not have the same dimensions, and are non-commuting, in general.
Edit: The matrices are non-commuting, in general.
*edit: I have to correct myself. I had incorrectly permuted terms (without noticing) which of course in general for matrices doesn't work.
What you can use is $(A \otimes B)\cdot(C \otimes D) = (AC \otimes BD)$. The latter implies $(A \otimes B)^n = A^n \otimes B^n$. However, expanding the $N$-th power of a sum is in general a bit more tedious for matrices. You have $$(P+Q)^2 = P^2 + PQ + QP + Q^2,$$ which for you becomes $$A^2\otimes B^2 + A(1-A) \otimes BD + (1-A) A \otimes DB + (1-A)^2 \otimes D^2.$$ Likewise, for $N=3$ you would need $$(P+Q)^3 = P^3 + PPQ + PQP + QPP + PQQ + QPQ + QQP + Q^3,$$ which would give you $$A^3\otimes B^3 + A^2(1-A)\otimes B^2D + A(1-A)A \otimes BDB + (1-A) A^2 \otimes DB^2 + A(1-A)^2 \otimes BD^2 + (1-A) A (1-A) \otimes DBD + (1-A)^2 A \otimes D^2B + (1-A)^3 \otimes D^3,$$ i.e., just all combinations, leading to $N^3$ terms.
Now, what you can do is use the fact that $A$ and $1-A$ commute. This lets you write $$A^2\otimes B^2 + A(1-A) \otimes (BD+DB) + (1-A)^2 \otimes D^2.$$ for $N=2$ and $$A^3\otimes B^3 + A^2(1-A)\otimes (B^2D+BDB + DB^2) + A(1-A)^2 \otimes (BD^2 + DBD + D^2B)+ (1-A)^3 \otimes D^3$$ for $N=3$, respectively.
The answer I was originally posting only works if $B$ and $D$ were to commute. In this case you had $$[A \otimes B + (1-A) \otimes D)^N = \sum_{n=0}^N {N \choose n} \Big(A^n (1-A)^{N-n} \otimes B^n D^{N-n}\Big).$$ but that doesn't help you as yours don't.
You do have a sum over $A^n (1-A)^{N-n}$ but each term is Kronecker-ed with another sum that considers all ${N \choose n}$ arrangements of $n$ times the $B$ matrix and $N-n$ times the $D$ matrix.
