Looks to me like this is false. Let $K = \mathbb{Q}(z)/(z^p-1-p^2)$. This is an extension of degree $p$, so it is disjoint from the p-th cycloctomic field, and hence does not contain a $p$-th root of $1$. Thus, it also can not contain a $p^k$-th root of 1.
Now, let's see how $z^p - 1 - p^2$ factors in Qp. There is already one p-th root of $1+p^2$ in Qp; call this root a. (To see this, note that the power series $(1+x)^{1/p} = 1+(1/p) x + \binom{1/p}{2} x^2 + ...$ converges for $x=p^2$.)
Let $\mathcal{P}$ be a prime of $K$ corresponding to a factor of $z^p-1-p^2$ other than z-a. (In fact, $( z^p-1-p^2)/(z-a)$ is irreducible over $\mathbb{Q}_p$, but I don't need that.) So $K_\mathcal{P}$ contains a root b of $z^p-1-p^2$ other than $a$. But then $b/a$ is in $K_\mathcal{P}$ and is a $p$-th root of 1.
This might be true if you ask $K/\mathbb{Q}$ to be Galois, but I would bet against it.
Let $L/K$ be a finite abelian extension of local fields. Although, there is no generic form for the image of the norm map, $N^{L}_K$, in practice one can follow the following procedure to determine its image.
Choose a uniformizer $\pi_L$ in $\mathcal{O}_L.$ Then $L^{\times}$ is equal to the group generated by $\pi_L$ and $\mathcal{O}_L^{\times}.$ It follows that $N^{L}_K L^{\times} = \langle N^{L}_K(\pi_L) \rangle N^{L}_K \mathcal{O}_L^{\times},$ Hence to determine $N^{L}_K L^{\times}$ it is enough to establish the image of $\mathcal{O}_L^{\times}$ under the norm mapping.
The group $N^L_K \mathcal{O}_L^{\times}$ is a subgroup of $\mathcal{O}_K^{\times}$ and by a group cohomology argument it can be shown that
$[\mathcal{O}_K^{\times}:N^{L}_K\mathcal{O}_L^{\times}] = e(L|K) = [L^{\times} : K^{\times}\mathcal{O}_L^{\times}].$
In particular, if $L/K$ is unramified, $\mathcal{O}_K^{\times} =N^{L}_K\mathcal{O}_L^{\times}$ and hence $N^{L}_K L^{\times} = \langle N^{L}_K(\pi_L) \rangle \mathcal{O}_K^{\times}.$
The norm group of a tamely ramified extension is similarly easy to deduce. Write
$$\mathcal{O}_L^{\times} = \langle \zeta_{q_L - 1} \rangle U_L$$
where $q_L$ is the residue field characteristic of $L$ and $\zeta_n$ denotes a primitive n-th root of unity. Denote the residue field of $L$ by $l$ and that of $K$ by $k,$ then
$$N^L_K(\zeta_{q_L -1}) = N^l_k(\zeta_{q_L -1})^{e(L|K)} = \zeta_{q_K -1}^{e(L|K)}.$$
As $U_K$ is a pro-$p$ group and contains the image of $U_L$ under $N^L_K,$ we have in the tamely ramified case that $U_K = N^L_K(U_L)$ else $\mathcal{O}_K^{\times}/N^{L}_K\mathcal{O}_L^{\times}$ would contain an element of $p$-power order contradicting the equality
$$[\mathcal{O}_K^{\times}:N^{L}_K\mathcal{O}_L^{\times}] = e(L|K)$$
and the fact that $L/K$ was assumed tamely ramfied. It follows in the tamely ramified case that $$N^{L}_K L^{\times} = \langle N^{L}_K(\pi_L), \zeta_{q_K -1}^{e(L|K)}\rangle U_K$$
The case of wild ramification is more difficult. But two facts are helpful. First, in the case $K$ is a p-adic field $U_K^{ap^n} \supset 1 +\mathcal{M}_K^{2ne(k|\mathbb{Q}_p) + 1}$ where $a$ and $p$ are relatively prime. Hence, it is enough to determine the image of the norm mapping in the units of $\mathcal{O}_K \mod \mathcal{M}_K^{2ne(k|\mathbb{Q}_p) + 1},$ a finite set.
Another technique is to determine the higher ramification filtration of $Gal(L/K)^{ab}.$ In practice this can be done by examining the derivatives of the irreducible polynomial of $\pi_L.$ These groups map under the inverse of the artin map to the higher unit groups. As the domain of the inverse of the artin map is $K^{\times}/N^{L}_K L^{\times}$ their sizes reveal norm indexes within the higher unit groups. For a good exposition see Serre's book Local Fields.
Best Answer
In
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jmsj/1261734945
Chevalley shows the following related statement (Remarque p. 39): let $p$ be a prime, $K$ a field of characteristic different from $p$, and $\zeta$ a $p^e$-th primitive root of $1$ in some algebraic extension of $K$, where $e\geq 1$ is any integer. Assume moreover that $-1$ is a square in $K$ if $p=2$. Then if an element $x\in K(\zeta)$ is a $p^e$-th power, then it is already a $p^e$-th power in $K$.