Here, We are using chaining as a resolution technique so if a collision happens at some place then we will get an extra bucket at that place.
Now here in question, we have N slots and we don't want any insertion in the last K slots, so remaining slots where we can insert value is N-K slots.
First Insertion :
The probability that the first insertion done at N-K slots is, (N-K/N)
Second Insertion :
The probability that the second insertion also done at N-K slots is, (N-K/N) * (N-K/N) = (N-K/N)^2
Third Insertion :
The probability that the second insertion also done at N-K slots is, (N-K/N) * (N-K/N) * (N-K/N)= (N-K/N)^3
so for, rth Insertion :
The probability that the second insertion also done at N-K slots is, (N-K/N) * (N-K/N) * ...... r times= (N-K/N)^r
So answer is (N-K/N)^r which is nothing but (1-(K/N))^r Option B