$$P = \sum_{i=1,3,\dots,2k-1} i$$
transform the summation by putting $i = 2r-1 $
where lower bound is $2r-1=1 \rightarrow r=1$
and upper bound is $2r-1=2k-1 \rightarrow r=k$
so we get
$$P = \sum_{r=1,2\dots,k} 2r-1$$
similarly
$$Q = \sum_{i=2,4,\dots,2k} i$$
transform the summation by putting $i = 2r$
where lower bound is $2r=2 \rightarrow r=1$
and upper bound is $2r=2k \rightarrow r=k$
so we get
$$Q = \sum_{r=1,2\dots,k} 2r$$
doing $$Q – P = \sum_{r=1,2\dots,k} 2r – (2r-1) = \sum_{r=1,2\dots,k}1 = k$$
so ans is A i.e $P = Q – k$