You could use the chains, but I think it's more straightforward to use the original set up to look at what rules force doctors in S, and avoid those sufficient conditions.
To start with, I'd look at the rules that say J(r)
O (s) and P(r)
O(s) and K(s)
Of those two, I'd start with seeing what happens if I put J(s) or P(s) so as not to trigger either sufficient condition. We know if J(s)
K(r). We know if P(s)....well, we don't know anything. It doesn't force anyone else in (s) either.
From there, we don't have any more rules that have (s) as a requirement. So we just have to look at the two rules we have, and see how to minimize the number of doctors in (s).
No matter where we put J, someone is in (s). Either J is in r, putting O in s, or we put J directly in s. That's one doctor.
For P, we could have either one or two doctors in s, depending on where P is placed. If we place P in r, two doctors (O and K) are in s. If we place P in s, one doctor (P) is added to s.
So no matter what, our minimum here is going to be two doctors in s---J and P. You might think that the Os overlap, and you could do something with those, but as it turns out, if N is in r, O has to be as well, so we want to avoid a situation with O in s. Otherwise we would have to bring N along with O to s, and we'd have three doctors there.
In summary, our minimum is 2. We can't get any lower than that under the rules.
Hope that helps.