Couple of ways of doing this one:

Let the couples be A1A2 and B1B2. Let the fifth wheel (the single person) be C.
We know that without any restrictions we can arrange the five people (A1)(A2)(B1)(B2)(C) in 5! ways.
Hence total ways = 5! = 120.

1st way: Count the unfavorable cases

Then:
a. No. of ways of A1A2 sitting together, (A1A2)(B1)(B2)(C) = 4!*2! = 48
b. Same way the number of ways of B1B2 sitting together (B1B2)(A1)(A2)(C) = 4!*2! = 48
c. We have counted the ways in which A1A2 and B1B2 both sit together twice (first as a part of a. and then as a part of b.) So, we need to remove one of the cases. No of ways of A1A2, B1B2 and C sitting together (A1A2)(B1B2)(C) = 3!*2!*2! = 24

So the total no of unfavorable ways = a+b-c = 48+48-24 = 72.
Total number of favorable ways = 120 - 72 = 48.
Then probability required (favorable) = 48/120 = 2/5.

2nd way: Count the number of favorable cases (using fundamental principle of counting).

Consider 5 chairs on which you have to sit 5 people down. _ _ _ _ _. Let's count the favorable cases as the single person (C) moves.

a. Single person at first chair C _ _ _ _ . Since first person is fixed, for the second chair, we have 4 choices. For the third, once the second is selected, we have 2 choices (since after selecting the person for the second chair, the third one can't have the person's partner). For the fourth, we have only 1 choice and same for the fifth. So we have C 4 2 1 1. No of ways = 4*2*1*1 = 8.

In the same way, if the single person is at the last place (_ _ _ _ C), no. of ways = 8 as well.
Total ways when single person is at the extreme chairs = 8*2 = 16.

b. Single person sits on second chair. _ C _ _ _. In this case, we have 4 choices for the first chair. For the third chair, we again have 2 choices. (If you thought the third chair has 3 choices, it is not right because let's say if A1 is selected at first and A2 at third, it forces the last two chairs to be filled with B1 and B2, which is not a favorable case. Hence we have 4 C 2 1 1, in this case. So, number of ways with C at second chair = 4*2*1*1=8 . In the same way, when C is at the 4th position, no. of ways = 8. So total ways when C is at 2nd or 4th = 2*8 = 16.

c. Single person sits on the middle chair. _ _ C _ _ . In this case, we have 4 choices for the first chair, 2 choices for the second. But in this case, we also have two choices at the 4th chair, since the last two people can be seated either way. So we have 4 2 C 2 1. No. of ways when C is in the middle = 4*2*2*1 = 16.

So the total number of favorable cases = 16+16+16. Hence probability required = 48/120 = 2/5.

So D is the right option.

Let me know if this helps