You should have what you need to make this happen.
I think there’s something funky with sqrt(nu / u), cause if that’s a vector and L * z is a vector, I don’t see how they multiply together. Assuming that sqrt(nu / u) stuff can get crammed in L, call that \hat{L}.
Anyway, so you can evaluate p(z), but what you have is x and a 1-to-1 transformation between x and z. In terms of the multivariate change of variables in the manual (pg. 403 of 2.17) that is:
x = \mu + \hat{L} z = f(z)
p_x(x) = p_z(f^{-1}(x)) | \text{det} J_{f^{-1}}(x)|
Then
z = \hat{L}^{-1}(x - \mu) = f^{-1}(x)
J_{f^{-1}}(x) = \hat{L}^{-1}
\text{det} \hat{L}^{-1} = \frac{1}{\text{det} \hat{L}}
And the determinant of a lower triangular matrix is just the product of the diagonal elements.