I would agree with you that the cached answer looks wrong and that there is a missing factor of 15 in $p_2$ (and of 20 in $p_3$).
When you ask "What is the probability that 2 couples sit next to each other?" you are looking at what was originally the $p_2$ part of the "What is the probability that at least one of the wives ends up sitting next to her husband?" question.
$p_2$ is the probability that two particular couples sit with husband next to wife. As you say, the ${4 \choose 2} = 6$ for which two couples these are is included in the $4p_1-6p_2+4p_3-p_4$ formula.
I would then say that the two particular couples have each become one unit (with $(2!)^2$ ways of arranging the individuals within the couple), reducing the number of orderable units from $8$ to $6$ so we should have
$$p_2 = \frac{(2!)^2 \; 6!}{8!} \approx 0.071428571 $$
and similarly $p_1 = \frac{(2!)^1 \; 7!}{8!} = 0.25$, $p_3 = \frac{(2!)^3 \; 5!}{8!} \approx 0.023810$, and $p_4 = \frac{(2!)^4 \; 4!}{8!} \approx 0.00952381$.
For $p_2$ and $p_3$, these are not the same as the cached answer, though the $p_2$ is the same as yours. Using the inclusion-exclusion formula this gives me a probability that at at least one of the wives ends up sitting next to her husband of about $0.657142857$, not the $0.9666667$ of the cached answer.
Some simulation convinces me this looks plausible.
First, judging from your work, you misstated the question: it appears that you meant to ask for the probability that no husband sits next to his wife. It also appears that the seats are arranged in a row, not a circle. I will make these assumptions.
Your first three factors, $8\cdot6\cdot5$, in the numerator are fine. After that, though, you have to split the calculation into cases. Suppose that the first three people, in order, are A, B, and C. If A and C are a couple, then any of the remaining $5$ people can sit in the fourth seat. Your calculation, with a factor of $4$, is correct only if C’s spouse has not already been seated, i.e., C and A are not a couple. And the cases just proliferate after that, which is why you’re better off working with the complement.
Best Answer
Think of the couple as if they were one person, then you have $9!$ ways for them to seat, but of course the couple can always switch seats so at the end you get $2\times9!$ ways.
So the probability that they sit together is $\frac{2.9!}{10!}=0,2$