Local Artin reciprocity, for a finite abelian extension $L/K$ with $K$ a $p$-adic field, is a specific isomorphism $K^{\times}/N_{L/K}(L^{\times}) \cong Gal(L/K)$. You seem particular interested in the unramified case, but let me treat the arbitrary case first.
Passing to the inverse limit over $L$, one gets an isomorphism $\varprojlim{} K^{\times}/N_{L/K}(L^{\times}) \cong G_K^{ab}.$ As Mephisto notes in their answer, the norm groups range over all open subgroups of $K^{\times}$, and so we may rewrite this as an isomorphism $\widehat{K^{\times}} \cong G_K^{ab},$ where $\widehat{K^{\times}}$ is the profinite completion of $K^{\times}$.
Recall that, if we choose a uniformizer $\pi$ for $K$, then $K^{\times} \cong \mathcal O^{\times} \times \mathbb Z$, where $\mathcal O$ denotes the ring of integers in $K$, and the isomorphism is given by mapping an element $a \in K^{\times}$ to $\bigl(a/\pi^{v(a)}, v(a) \bigr),$ where $v: K^{\times} \to \mathbb Z$ is the valuation, normalized via $v(\pi) = 1$.
Thus $\widehat{K^{\times}} \cong \mathcal O^{\times} \times \hat{\mathbb Z}$. (Recall that $\mathcal O^{\times}$ is its own profinite completion, but $\mathbb Z$ is not; we let $\hat{\mathbb Z}$ denote the profinite completion of $\mathbb Z$.)
Now we can understand what happens if we restrict to unramified extensions. The valuation $v: K^{\times} \to \mathbb Z$ induces a projection $\widehat{K^{\times}} \to \hat{\mathbb Z}$, which is independent of the choice of $\pi$. Similarly, there is a surjection $G_K^{ab} \to Gal(K^{nr}/K)$. The latter group is isomorphic to $\widehat{Z}$; it is naturally identified with the absolute Galois group of the residue field, and is topologically generated by Frobenius.
Under the reciprocity isomorphism $\widehat{K^{\times}} \cong G_K^{ab}$, the projection to $\widehat{Z}$ on the source and the projection to $Gal(K^{nr}/K)$ on the target are compatible with the isomorphism $\widehat{Z} \cong Gal(K^{nr}/K)$ given by mapping $1$ to Frobenius.
Now we can bring in Weil groups. The general definition of the Weil group is somewhat involved, involving the fundamental classes in $H^2(L/K, L^{\times})$, but after you sort everything out, you find that the Weil group $W_K$ can be identified with a subgroup of $G_K$, namely the preimage under the natural map $G_K \to Gal(K^{nr}/K) \cong \hat{\mathbb Z}$ of the subgroup $\mathbb Z \subset \widehat{\mathbb Z}$. From this, one sees that $W_K^{ab}$ can be identified with the subgroup of $G_K^{ab}$ which again is the preimage under the natural map $G_K \to Gal(K^{nr}/K) \cong \hat{\mathbb Z}$ of the subgroup $\mathbb Z\subset \widehat{\mathbb Z}$.
If we go back to the preceding discussion of our reciprocity map, we see that the isomorphism $\widehat{K^{\times}} \cong G_K^{ab}$ restricts to an isomorphism $K^{\times} \cong W_K^{ab}$. The inverse of this is the isomorphism you are looking for, I would guess.
If one restricts to the unramified case, then we have to pass to the quotient $\mathbb Z$ of $K^{\times}$, and the quotient $\mathbb Z \subset \widehat{\mathbb Z} = Gal(K^{nr}/K)$ of $W_K^{ab}$, and then the isomorphism just becomes $\mathbb Z = \mathbb Z$.
Additional remarks: There are various confusions in your question. Here are some: you write $W_K$ but you mean $W_K^{ab}$. (The Weil group itself is not abelian, just as $G_K$ is not abelian.) You say you want the general Artin reciprocity isomorphism $W_K$ [sic] $\to K^{\times}$, but you then restrict attention to unramified extensions. As noted above, these will only see the quotient $\mathbb Z$ of $W_K^{ab}$ and the corresponding quotient $\mathbb Z$ of $K^{\times}$. You then go on to ask for a continuous homomorphism $\varprojlim{} K^{\times}/N_{L/K}(L^{\times}) \to K^{\times}.$ Leaving aside the issue that the left-hand side should involve all abelian $L$ over $K$, not just the unramified ones (or else the left-hand side will be too small), there is no such map, since the left-hand side is profinite and the right hand side has a discrete factor (as noted above). The correct thing is to profinitely complete $K^{\times}$, and one then has the reciprocity isomorphism discussed above.
Further remarks: Here are some additional remarks, prompted in part by the exchange of comments below.
At a technical level, passing from $G_K$ to $W_K$ simply replaces the $\widehat{\mathbb Z}$ quotient of $G_K$ coming from the map $G_K \to Gal(K^{nr}/K) \cong \widehat{\mathbb Z}$ by $\mathbb Z$, so that instead of writing the reciprocity isomorphism as an isomorphism $\widehat{K^{\times}} \cong G_K^{ab}$ we can rewrite it as an isomorphism $K^{\times} \cong W_K^{ab}$, and so avoid passing from $K^{\times}$ to $\widehat{K^{\times}}$.
As for why we do this: one reason is that $K^{\times}$ is what appears as a local factor in the ideles, not $\widehat{K^{\times}}$. There are additional motivations. One thing to remember is that the original definition of the Weil group is not as the preimage of $\mathbb Z \subset \widehat{\mathbb Z}$ in $G_K$, but in terms of the fundamental classes of local class field theory; so the Weil group arises naturally from this point of view. For further motivations, this MO post might help.