How does this resolve? Supposedly the conclusion follows but I don't see how, since if we imagine $S$ as false but $P$ as true, then things seem to check out. (Classical predicate calculus, use elementary inference rules in a discrete math course, for example).

HYP 1: $P\vee \neg Q$

HYP 2: $Q \vee R$

HYP 3: $\neg R \vee S$

HYP 4: $P$

Thus $S$ 

Thanks, sorry for missing proper inference symbols. My professor is insistent that this works but I used contradiction and there was no contradiction.
I hope this can help:

p ∧(p ∨ ⊣q)⋀(q⋁r)⋀(⊣r⋁s)

≡p∧(q∨r)∧(⊣r⋁s)

Since p is always true, according to hypothesis we have:

((q∨r)∧⊣r)∨((q∨r)∧s)

≡((q∧⊣r)⋀F)⋁((q⋀s)⋁(r⋀s))

≡(q∧⊣r)⋁(q⋀s)⋁(r⋀s)

≡(q∧(⊣r⋁s))⋁(r⋀s)

As mentioned in the problem, ⊣r⋁s is always true. Now we have:

q∨(r∧s)≡T

(q∨r)∧(q∨s)≡T

And we know q∨r is always true, so:

q∨s≡T

This shows that we can conclude (q or s is true) and we can't conclude that s is true.

This is what I think about this problem, I hope I have solved it truly!!!
I hope this can help:

p ∧(p ∨ ⊣q)⋀(q⋁r)⋀(⊣r⋁s)

≡p∧(q∨r)∧(⊣r⋁s)

Since p is always true, according to hypothesis we have:

((q∨r)∧⊣r)∨((q∨r)∧s)

≡((q∧⊣r)⋀F)⋁((q⋀s)⋁(r⋀s))

≡(q∧⊣r)⋁(q⋀s)⋁(r⋀s)

≡(q∧(⊣r⋁s))⋁(r⋀s)

As mentioned in the problem, ⊣r⋁s is always true. Now we have:

q∨(r∧s)≡T

(q∨r)∧(q∨s)≡T

And we know q∨r is always true, so:

q∨s≡T

This shows that we can conclude (q or s is true) and we can't conclude that s is true.

This is what I think about this problem, I hope I have solved it truly!!!