Let us assume that $A-B$ is invertible. Write

$$\begin{array}{rl} C &= \begin{bmatrix} A & B & \ldots & B\\ B & A & \ldots & B\\ \vdots & \vdots & \ddots & \vdots\\B & B & \ldots & A\end{bmatrix}\\\\ &= \begin{bmatrix} A-B & O_n & \ldots & O_n\\ O_n & A-B & \ldots & O_n\\ \vdots & \vdots & \ddots & \vdots\\ O_n & O_n & \ldots & A-B \end{bmatrix} + \begin{bmatrix} B \\ B\\ \vdots \\ B\end{bmatrix} \begin{bmatrix} I_n \\ I_n\\ \vdots \\ I_n\end{bmatrix}^T\\\\ &= (I_k \otimes (A-B)) + (1_k \otimes A) (1_k \otimes I_n)^T\\\\ &= (I_k \otimes (A-B)) \left(I_{nk} + (I_k \otimes (A-B)^{-1}) (1_k \otimes B) (1_k \otimes I_n)^T\right)\end{array}$$

Using Sylvester's determinant identity,

$$\begin{array}{rl} \det (C) &= \det\left((I_k \otimes (A-B)) \left(I_{nk} + (I_k \otimes (A-B)^{-1}) (1_k \otimes A) (1_k \otimes I_n)^T\right)\right)\\\\ &= \det(I_k \otimes (A-B)) \cdot \det \left( I_{nk} + (I_k \otimes (A-B)^{-1}) (1_k \otimes B) (1_k \otimes I_n)^T \right)\\\\ &= \det(I_k \otimes (A-B)) \cdot \det \left( I_{n} + (1_k \otimes I_n)^T (I_k \otimes (A-B)^{-1}) (1_k \otimes B) \right)\\\\ &= \left(\det(A - B)\right)^k \cdot \det \left( I_{n} + k (A-B)^{-1} B \right)\\\\ &= \det((A-B)^k) \cdot \det \left( I_{n} + k (A-B)^{-1} B \right)\\\\ &= \det((A-B)^{k-1}) \cdot \det \left( A-B + k B \right)\\\\ &= (\det(A-B))^{k-1} \cdot \det \left( A + (k-1) B \right)\end{array}$$

Let us assume that $A-B$ is invertible. Write

$$\begin{array}{rl} C &= \begin{bmatrix} A & B & \ldots & B\\ B & A & \ldots & B\\ \vdots & \vdots & \ddots & \vdots\\B & B & \ldots & A\end{bmatrix}\\\\ &= \begin{bmatrix} A-B & O_n & \ldots & O_n\\ O_n & A-B & \ldots & O_n\\ \vdots & \vdots & \ddots & \vdots\\ O_n & O_n & \ldots & A-B \end{bmatrix} + \begin{bmatrix} B \\ B\\ \vdots \\ B\end{bmatrix} \begin{bmatrix} I_n \\ I_n\\ \vdots \\ I_n\end{bmatrix}^T\\\\ &= (I_k \otimes (A-B)) + (1_k \otimes B) (1_k \otimes I_n)^T\\\\ &= (I_k \otimes (A-B)) \left(I_{nk} + (I_k \otimes (A-B)^{-1}) (1_k \otimes B) (1_k \otimes I_n)^T\right)\end{array}$$

Using Sylvester's determinant identity,

$$\begin{array}{rl} \det (C) &= \det\left((I_k \otimes (A-B)) \left(I_{nk} + (I_k \otimes (A-B)^{-1}) (1_k \otimes B) (1_k \otimes I_n)^T\right)\right)\\\\ &= \det(I_k \otimes (A-B)) \cdot \det \left( I_{nk} + (I_k \otimes (A-B)^{-1}) (1_k \otimes B) (1_k \otimes I_n)^T \right)\\\\ &= \det(I_k \otimes (A-B)) \cdot \det \left( I_{n} + (1_k \otimes I_n)^T (I_k \otimes (A-B)^{-1}) (1_k \otimes B) \right)\\\\ &= \left(\det(A - B)\right)^k \cdot \det \left( I_{n} + k (A-B)^{-1} B \right)\\\\ &= \det((A-B)^k) \cdot \det \left( I_{n} + k (A-B)^{-1} B \right)\\\\ &= \det((A-B)^{k-1}) \cdot \det \left( A-B + k B \right)\\\\ &= (\det(A-B))^{k-1} \cdot \det \left( A + (k-1) B \right)\end{array}$$