A set is countable means there exist a enumeration procedure to generate each of its elements and for a given element of set, it take finite step to generate it using the enumeration procedure.
Let $\Sigma = \{a,b \}$ and there exist a enumeration procedure to generate all the string of the language $\Sigma^*$.
$\Sigma^{*}=\{ \epsilon , a , b , aa, ab, ba, bb , aaa , \ldots \}$
Here, enumeration procedure is simply the generating string of the language by length for the fixed length string are in alphabetical order.
This way $\Sigma^{*}$ is countably infinite & $2^{\Sigma ^{*}}$ will be uncountable set
Because the power set of any countably infinite set is uncountable.
Ref: http://www.cs.xu.edu/csci250/06s/Theorems/powerSetuncountable.pdf
Correct Answer: $C$