We have $n$ Kingdoms as ${k_1}, {k_2},\ldots , k_n.$
Firstly we can select $2$ champions from ${2n}$ champions and assign to ${k_1}= \binom{2n}{2}$ ways (Say ${w_1}$)
Then we can select next $2$ champions and assign to ${k_2} = \binom{2n-2}{2}$ ways (Say ${w_2}$)
and so on..
For last kingdom , we have $2$ champions left$ = \binom{2}{2}$ ways (Say ${w_n}$)
$\begin{align}\text{Total ways for assigning $2n$ champions to $n$ kingdoms} &= {w_1} * {w_2} *\ldots *{w_n} \\ &= \binom{2n}{2} * \binom{2n - 2}{2} * . . . * \binom{2}{2} \\&=\frac{(2n)!}{2^{n}} \end{align}$
So, Option A (Ans) .