Option $(B)$ is the correct answer.
$P(H(L) > H(K)) + P(H(K) > P(L)) + P(H(K) == H(K)) = 1$
Now since, $P(H(L) > H(K))$ and $P(H(K) > P(L))$ are symmetrical, they will be equal (lets say = $Y$).
Assume, $P(H(K) == H(K)) = X$
So, $X + 2*Y = 1$
$\Rightarrow Y = (1-X)/2$
And, $X = P(H(L)==0$ AND $H(K)==0)$ + $P(H(L)==1$ AND $H(K)==1)$ + .... + $P(H(L)==n$ AND $H(K)==n)$
$X = \binom{n}{0}^2/2^{2n} + \binom{n}{1}^2/2^{2n} + .... + \binom{n}{n}^2/2^{2n}$
$X = \sum_{i=0}^{n} \binom{n}{i}^2/2^{2n}$
$Y = \frac{1}{2}(1 - \sum_{i=0}^{n} \binom{n}{i}^2/2^{2n})$