It immediately follows from the monotonicity property that,
$0\leq P(E)\leq 1,$
Probability of at least one means union of the probability of events, i.e.,
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$
here, $P(A\cup B) = 1$ , because it can not be more than $1$ and if at least one of the event has probability $1$ (here , $P(A) = 1$), then union of both should be $1.$
So,
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$
$1 = 1 + \dfrac{1}{2} - P(A\cap B) ,$
$P(A\cap B) =\dfrac{1}{2},$
Now,
$P(A\mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{\left(\dfrac{1}{2}\right)}{\left(\dfrac{1}{2}\right)} = 1 ,$
$P(B\mid A) = \dfrac{P(A\cap B)}{P(A)} =\dfrac{\left(\dfrac{1}{2}\right)} { 1} =\dfrac{1}{2} .$
Hence, option is $(D)$.
NOTE :- if at least one of the two events has probability 1, then both events should be independent but vise versa is not true.