You can easily understand by this example:
let R(x,y) be x is greater than y.
then R(y,x) will be y is greater than x.
now for all x for all y (R(x,y) -> R(y,x))
can be proved to be false if we can come up with the case as "T->F". This can be easily done by taking value of x as 6 and value of y as 2.
so 6>2 is true and 2>6 is false. Therefore the value is false.
Since we have proved for some variables that the implied predicate is giving us false, therefore the given statement is definitely not valid.
But it is also not satisfiable till now. To make it satisfiable we must find at least one case for which we can get result as true.
This is also simple as take x = 2 and y = 6, so 2>6 is false and F implied anything is true. Therefore with x = 2 and y = 6 we made our predicate as True.
Since we got a true case therefore the predicate is "Satisfiable"
Now let's check for the negation.
the negation of predicate will be,
there exist x there exist y (R(x,y) ^ not of R(y,x))
not of R(y,x) means y is not greater than x.
let's take again x = 2 and y = 6, now this will make predicate false as F^anything is false.
therefore clearly negation of the given predicate is definitely not valid.
can we make it true for atleast a case?
let's check:
take x = 6 and y = 2
now 6>2 and also 2 is not greater than 6
therefore the given predicate is true.
this means we have atleast a case for which negation is true. Therefore the negation is satisfiable.
so the answer is option b. "satisfiable and so is it's negation".