I am unsure as to how to obtain the correct answer

Calculate [H+] and the % dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2x10^-4) and 1.0 M NaF?
do I make an ICE table with:

HF + NaOH <-> H2O + NaF
1 1
-x +x

It just doesn't make sense to me. I think I have the wrong initial equation. I know how to do % dissociation. I just can't get started. Any help please2

NaF is a strong electrolyte. So assume that its dissocation goes to completion and you'll have [F-]= 1 M.

The equlibrium equation is
HF <-> H+ + F-
I 1.0 0 1.0
C -x x +x
E 1-x x 1+x

Then use definition of K.
K = (1+x)(x)/(1-x).
Solve for x. That's your answer !