Answer:
b; ∀x P(x) ∨ ¬P(x)
Step-by-step explanation:
Suppose that we have a proposition p
Such that p can be true or false.
We can define the negation of p as:
¬p
Such that, if p is false, then ¬p is true
if p is true, then ¬p is false.
Also remember that a proposition like:
p ∨ q
is true when, at least one, p or q, is true.
Then if we write:
p ∨ ¬p
Always one of these will be true (and the other false)
Then the statement is true.
And if the statement depends on some variable, then we will have that:
p(x) ∨ ¬p(x)
is true for all the allowed values of x.
from this, we can conclude that the statement that is always true is:
b; ∀x P(x) ∨ ¬P(x)
Where here we have:
For all the values of x, P(x) ∨ ¬P(x)
. (Integration along the 
axis would be way easier, but whatever)