Answer: Option D
Let S be the sample space
$$\eqalign{
& {\text{Then}},n\left( S \right) = {}^{52}{C_2} \cr
& = \frac{{ {52 \times 51} }}{{\left( {2 \times 1} \right)}} \cr
& = 1326 \cr} $$
Let E = event of getting 2 kings out of 4
$$\eqalign{
& \therefore n\left( E \right) = {}^4{C_2} = \frac{{ {4 \times 3} }}{{ {2 \times 1} }} = 6 \cr
& \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr
& = \frac{6}{{1326}} \cr
& = \frac{1}{{221}} \cr} $$