The book asserts that $\Psi$ is a positive definite matrix and expresses it as $\Psi^{-1} = P'P$, where $P$ is a square non-singular matrix. The implication is that, due to the positive definiteness of $\Psi$, there always exists a matrix $P$ such that $\Psi^{-1} = P'P$.
However, based on my understanding, if $\Psi$ is a real positive definite matrix, the Cholesky decomposition can be employed to decompose it as $\Psi = PP'$, where $P$ is a real lower triangular matrix with positive diagonal entries. Consequently, the inverse of $\Psi$ can be expressed as $\Psi^{-1} = (P')^{-1}P^{-1}$.
What do I miss? The book states $\Psi^{-1} = P'P$, while the Cholesky decomposition suggests $\Psi^{-1} = (P')^{-1}P^{-1}$.
Your book says $\Psi^{-1} = S^*S$. Cholesky says $\Psi^{-1} = (L^*)^{-1}L^{-1}$. Let $S = L^{-1}$ and you are done. However, this decomposition is far from unique.
One could also obtain a factorization of this type from the spectral decomposition of $\Psi$. Since $\Psi$ is PSD, it has a unitary diagonalizing transformation $\Psi = U\Lambda U^*$. From positivity, $\Lambda^{1/2}$ is well-defined, so let $P = \Lambda^{1/2}U^*$ and $\Psi = P^*P$, as desired.