There is a correspondence between $k$-isomorphic finite separable normal extensions of $k$, $char(k) = p > 0$, and abelian p-groups.
Put it formally, $$\forall \, G = Gal(\Bbb F/k), |G| < \infty, \, char(k) = p > 0, \, \exists \, P \text{ - abelian p-group}, \exists \, \beta \text{ - bijection}: \beta(G) = P$$ up to $k$-isomorphism, and vise versa.
How to clearly define such a bijection (i.e. how to prove this)?
Well, since galois extension can be defined as automorphisms that fixes field $k$, and since $k$ has characteristic $p$, we can consider Frobenius endomorphism $\mathrm f: q \mapsto q^p$, which can be connected with p-groups as we talk about finite extensions (i.e. finite groups). $k$ is fixed by $G$, hence so do $\mathrm f(k), \mathrm f(\mathrm f(k))$ and so on. But one can generate Galois extension by Frobenius automorphisms only in case of finite fields, so the way is not so trivial.
It's not entirely clear what precise statement you are asking to be true (what you have written does not make sense if taken literally), but any reasonable interpretation is false and I don't know why you would think it is true. For instance, if $k$ is algebraically closed, then there is only one finite extension of $k$. Or if $k$ is uncountable, there could be uncountably many non-isomorphic finite Galois extensions, whereas there are only countably many finite abelian $p$-groups.