Let me assume that $\mathbf{J} \in \mathbb{R}^{m \times n},~m<n$ is a full row rank matrix, $\mathbf{A} \in \mathbb{R}^{n \times n}$ is a symmetric positive definite matrix, and $\mathbf{J}^{-}$ is a weighted right pseudoinverse of the matrix $\mathbf{J}$ given as $\mathbf{J}^{-}=\mathbf{A}^{-1}\mathbf{J}^{T}(\mathbf{J}\mathbf{A}^{-1}\mathbf{J}^{T})^{-1}$.
It seems to me that $\mathbf{J}\mathbf{A}^{-1}\mathbf{J}^{-}$ is a regular matrix, but I have problems to prove it. I would be grateful if somebody has an idea.
If you mean invertible by regular, then the answer is only if A is full-rank. The weighted pseudo-inverse formulation that you brought is only valid if A is full-rank. Otherwise, you can use a regularized pseudo-inverse (or damped least-square) $J^-=A^{-1}J^T(JA^{−1}J^T+\mu^2I)^{−1}$ where $\mu$ is a small number. Here is a more complete way of handling it: Weighted pseudo-inverse formulation