$k\log k = \Theta \left ( n \right )$
$\Rightarrow c1\cdot n \leq k\log k \leq c2\cdot n$
Now assuming $n$ and $k$ are greater than 1, taking $log$ of each of the terms above will not change the inequality so we can write,
$ \log \left ( c1 \cdot n \right ) \leq \log\left ( k \cdot \log k \right ) \leq \log\left ( c2\cdot n \right )$
$\Rightarrow \left ( \log \left ( c1 \right ) + \log\left ( n \right ) \right ) \leq \left ( \log\left ( k \right ) + \log\left ( \log k \right ) \right ) \leq \left ( \log\left ( c2 \right ) + \log\left ( n \right ) \right )$
Now since we know that
$\log\left ( \log k \right ) \leq \log\left ( k \right )$
let us write $\left ( \log\left ( k \right ) + \log\left ( \log k \right ) \right ) = 2\log\left ( k \right )$.
To maintain the inequality,
$\left ( \log\left ( k \right ) + \log\left ( \log k \right ) \right ) \leq \left ( \log\left ( c2 \right ) + \log\left ( n \right ) \right )$,
after changing $\left ( \log\left ( k \right ) + \log\left ( \log k \right ) \right )$ to $2\log\left ( k \right )$,
we should multiply $\left ( \log\left ( c2 \right ) + \log\left ( n \right ) \right )$ by $2$, to compensate the increment in the middle term.
$\Rightarrow \left ( \log \left ( c1 \right ) + \log\left ( n \right ) \right ) \leq 2\log\left ( k \right ) \leq 2\left ( \log\left ( c2 \right ) + \log\left ( n \right ) \right )$
To make this inequality more clear let's write it using new constants as follows,
$c3 \left ( \log \left ( n \right ) \right ) \leq c4 \left ( \log \left ( k \right ) \right ) \leq c5 \left ( \log \left ( n \right ) \right )$
On dividing $ c1\cdot n \leq k\log k \leq c2\cdot n$ by the above inequality we get,
$\left ( \frac{c1}{c3} \right ) \left ( \frac{n}{\log\left ( n \right )} \right ) \leq \left ( \frac{1}{c4} \right )\left ( \frac{k\log k}{\log\left ( k \right )} \right ) \leq \left ( \frac{c2}{c5} \right )\left ( \frac{n}{\log\left ( n \right )} \right )$.
$\Rightarrow \left ( \frac{c1}{c3} \right ) \left ( \frac{n}{\log\left ( n \right )} \right ) \leq \left ( \frac{1}{c4} \right )k \leq \left ( \frac{c2}{c5} \right )\left ( \frac{n}{\log\left ( n \right )} \right )$
$\Rightarrow \left ( \frac{c1 \cdot c4}{c3} \right ) \left ( \frac{n}{\log\left ( n \right )} \right ) \leq k \leq \left ( \frac{c2 \cdot c4}{c5} \right )\left ( \frac{n}{\log\left ( n \right )} \right )$
$\Rightarrow k = \Theta \left ( \frac{n}{\log\left ( n \right )} \right )$.
PS - I am not very sure about the inequalities division step.
