I think writing things out in sums for this problem obscures what's really going on. I think keeping things as matrices and vectors makes them a lot clearer.
[tex] p({\bf x}) = (2\pi)^{N/2} {\bf C}^{1} \exp\left(\frac{1}{2}({\bf x}{\bf \mu})^T {\bf C}^{1}({\bf x}{\bf \mu})\right)[/tex]
[tex] p({\bf k}) = (2\pi)^{N/2} {\bf C}^{1} \int \exp\left(\frac{1}{2}({\bf x}{\bf \mu})^T {\bf C}^{1}({\bf x}{\bf \mu})+ i {\bf k}^T {\bf x}\right) d {\bf x}[/tex]
Now, you correctly notices that
[tex] {\bf C} = {\bf D}{\bf \sigma}{\bf D}^{1}[/tex]
Where [itex]{\bf \sigma}[/itex] is a diagonal matrix and [itex]{\bf D}[/itex] is orthogonal. However did you know that [itex]{\bf D}[/itex] being orthogonal means that
[tex]{\bf D}^{1} = {\bf D}^{T} [/tex]
Okay so define
[tex] {\bf y} = {\bf D}({\bf x}{\bf \mu})[/tex]
Hence
[tex] {\bf x} = {\bf D}^T{\bf y}+{\bf \mu}[/tex]
And so after a change of variables in the integral we get
[tex] p({\bf k}) = (2\pi)^{N/2} {\bf C}^{1} \int \exp\left(\frac{1}{2}{\bf y}^T {\bf \sigma}^{1}{\bf y}+ i {\bf k}^T ({\bf D}^T{\bf y}+{\bf \mu}))\right) d {\bf y}[/tex]
Pull out the mean
[tex] p({\bf k}) = (2\pi)^{N/2} {\bf C}^{1} \exp( i {\bf k}^T{\bf \mu})\int \exp\left(\frac{1}{2}{\bf y}^T {\bf \sigma}^{1}{\bf y}+ i {\bf k}^T{\bf D}^T{\bf y})\right) d {\bf y}[/tex]
Okay now hint: the [itex]{\bf D}[/itex] in the integral is a rotation. Think about it rotating in kspace as opposed to real space. So define
[tex] {\bf \kappa} = {\bf D} {\bf k} [/tex]
