It's Simple,
$ L = \{ w | n_a (w) = n_b (w) \} $
Its pushing $0$ for each $a$, and $1$ for each $b$. If its getting $b$ and at the top of stack we have $0$ then its performing pop operation. Similarly, If its getting $a$ and at the top of stack we have $1$ then its performing pop operation.
And Yes, If we make both the states final state then language will be $ {(a + b)}^* $.