Let $p$ be an odd prime. Then $\mathbb{Z}_p^{\times} \cong \mathbb{Z}/(p-1) \oplus \mathbb{Z}_p.$ The pro-$p$ sylow of $\mathbb{Z}_p^{\times}$ is $1 + p\mathbb{Z}_p$ and is isomorphic (as a topological group) to $p\mathbb{Z}_p$ via the $p$-adic logarithm. Since $1 + p^2\mathbb{Z}_p$ is a closed subgroup of $1 + p\mathbb{Z}_p$ of index $p$ it must be the case that $1 + p^2\mathbb{Z}_p$ is the pullback of $p^2\mathbb{Z}_p$ under the logarithm. From this it follows
$ 1 + p^2\mathbb{Z}_p = (1 + p\mathbb{Z}_p)^p.$
And therefore,
$(\mathbb{Z}_p^{\times})^p = \mu_{p-1}(1 + p^2\mathbb{Z}_p).$
Now consider an element $x \in \mathbb{Q}_p^{\times}$ and recall $x = ap^n$ for some $a\in\mathbb{Z}_p^{\times}$ and $n\in\mathbb{Z}.$ Then $x$ has a $p$-th root in $\mathbb{Q}_p$ if and only if $a$ has a root in $\mathbb{Z}_p^{\times}$ and $p|n.$
Hence,
$(\mathbb{Q}_p^{\times})^p = \mathbb{Z}_p^{\times}\langle p^p \rangle = \mu_{p-1}(1 + p^2\mathbb{Z}_p)\langle p^p \rangle.$
Now try to answer the question for $p =2.$