Let $\det(\cdot)$ denote the determinant and let $\operatorname{Det}(\cdot)$ denote the pseudo-determinant, which for any square matrix $A \in \mathbb{R}^{n \times n}$ can be defined as $$ \operatorname{Det}(A) := \lim_{\delta \to 0} \frac{\det(A + \delta I)}{\delta^{n - \operatorname{rank}(A)}}. $$
Now, let $A \in \mathbb{R}^{k \times n}$, $b \in \mathbb{R}^k$ be a vector with strictly positive entries, and assume that $k \geq n$. For the usual determinant, we have the identity $$ \det(A^T \operatorname{diag}(b) \, A) = \det(\operatorname{diag}(b)) \det(A A^T). $$ In the case that $\ker(A) \neq \{ 0 \}$, then both sides of the above reduce to 0.
Question: is there an analogous way to express the pseudo-determinant $$ \operatorname{Det}(A^T \operatorname{diag}(b) \,A) $$ as a product of factors involving determinants and/or pseudo-determinants of the matrices involved?
It is clear that in general $\operatorname{Det}(A^T \operatorname{diag}(b) \,A) \neq \det(\operatorname{diag}(b)) \operatorname{Det}(A A^T)$. I have tried manipulating the limit definition of the pseudo-determinant using the SVD of $A$ and the matrix determinant lemma, but I have not managed to get a clean expression out of this. I am beginning to think that there is no simple expression, or at least not one that is cleaner than given here.
For example, two exemplars I have computed are: $$ A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \quad \operatorname{Det}\left( A^T \operatorname{diag}(b) \, A \right) = 2(b_1 + b_2), \quad \operatorname{Det}\left( A A^T \right) = 4, $$ and $$ A = \begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}, \quad \operatorname{Det}\left( A^T \operatorname{diag}(b) \, A \right) = 3(b_1 b_2 + b_2 b_3 + b_1 b_3), \quad \operatorname{Det}(A A^T) = 9. $$