Let us Consider the options one by one :
(a) ((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q) This option is clearly true as this is well known rule for Resolution.
Consider option (c)
(P ∨ Q) is satisfiable if and only if (P ∨ R) ∧ (Q ∨ ¬R) is satisfiable
which means
(P ⋁ Q) ↔ (P V R) $\wedge$ (Q V ~R)
So it means (P V Q) is true if and only is (P V R) ^ (Q V ~R) is true
Let's suppose assume
(P V R) $\wedge$ (Q V ~R) is true
So Either of R or ~R can be true at a time.
Say, R is true so
(P V R) term becomes true
Now for the (P V R) $\wedge$ (Q V ~R) to become true
Q must be true
and If Q becomes true, then the LHS of bi-implication i.e. P V Q will become true hence making option (C) as true.
Now consider option (d)
(P ∨ Q) ⇒ FALSE if and only if both P and Q are unsatisfiable
Yes this is very obvious as when both P and Q are FALSE, P V Q will result in FALSE, making the above implication true.
Being P and Q unsatisfiable means they are FALSE or a Contradiction.
So, this option is also TRUE.
Remaining option is Option (b)
(P ∨ Q) ⇒ ((P ∨ R)) ∧ (Q ∨ ¬R)) is logically valid
For the above to be logically valid, it must be a tautology.
Let's check
(P+Q) -> ((P+R).(Q +~R))
~(P+Q) + (PQ + P~R + QR)
~P~Q + PQ + P~R + QR
and this is definitely not a tautology.
Hence, answer option (B)