To sort $k$ elements in a heap, complexity is $\Theta (k \log k)$. Lets assume there are $\frac{\log n}{\log \log n}$ elements in the heap.
Complexity $= \Theta\left(\frac{\log n}{\log \log n} \log\left(\frac{\log n}{\log \log n}\right)\right) $
$=\Theta \left( \frac{\log n}{\log \log n} \left({\log \log n}- { \log \log \log n}\right) \right )$
$= \Theta \left ({ \log n} - \frac{ \log n \log \log \log n}{\log \log n} \right )$
$=\Theta (\log n)$ (as shown below)
So, (C) is the answer.
$\log \log n > \log \log \log n$
$\implies \frac{\log \log \log n}{\log \log n} < 1$
$\implies \frac{ \log n \log \log \log n}{\log \log n} < \log n$
$\implies \Theta \left ( { \log n} - \frac{ \log n \log \log \log n}{\log \log n} \right ) =\Theta (\log n)$