Let $p \geq 2$ be prime and $K=\mathbb Q(\zeta_p),~\zeta^{p}=1$ with ring of integers $\mathcal{O}_K$. we denote by $\mathfrak{p} \mid p$ the prime ideal of $K$ dividing $p$. Let $K_{\mathfrak{p}}$ be the $\mathfrak{p}$-completion of $K$. We can say that $K_{\mathfrak{p}}$ is a cyclotomic extension of the $p$-adic field $\mathbb Q_p$. Denote $\mathbb Z_p$ be the ring of $p$-adic integers in $\mathbb Q_p$, let $\mathcal{O}_{K_{\mathfrak{p}}}$ be ring of integers of $K_{\mathfrak{p}}$ and $\mathfrak{q}$ be the corresponding prime-ideal (infact, maximal ideal) in $K_{\mathfrak{p}}$.
This post discusses about global units of a number field. Define the principal global units of $K$ by $$U_K=\{x \in \mathcal{O}_K^{\times}~:~x \equiv 1~(\text{mod}~\mathfrak{p})\}$$ and the principal local units of $K_{\mathfrak{p}}$ by $$U_{K_{\mathfrak{p}}}=\{x \in \mathcal{O}_{K_{\mathfrak{p}}}^{\times}~:~x \equiv 1~(\text{mod}~\mathfrak{q})\}.$$ From local-field theory, we can decompose $U_{K_{\mathfrak{p}}}$ as $$U_{K_{\mathfrak{p}}}=1+\mathfrak{q}=\mu_p \oplus (1+\mathfrak{q}^2),$$ where $\mu_p$ is the group of $p$-th roots of unity.
The above decomposition follows from $$1+\mathfrak{q}/1+\mathfrak{q}^2 \simeq \mathfrak{q}/\mathfrak{q}^2 \simeq \kappa_{K_{\mathfrak{q}}}.$$ My question:
Can we similarly decompose the principal global units $U_K=1+\mathfrak{p}$ as $1+\mathfrak{p}=\mu_p \oplus (1+\mathfrak{p}^2)$ ?
Does this holds for all primes $p$ ?
Note that $\mathfrak{p} \subset \mathfrak{q} \Rightarrow \mathfrak{p}^2 \subset \mathfrak{q}^2$, which implies $1+\mathfrak{p} \subset 1+\mathfrak{q}$ as well as $1+\mathfrak{p}^2 \subset 1+\mathfrak{q}^2$. Therefore, \begin{align} &1+\mathfrak{p}/1+\mathfrak{p}^2 \subset 1+\mathfrak{q}/1+\mathfrak{q}^2 \simeq \kappa_{K_{\mathfrak{p}}} \\ &\Rightarrow 1+\mathfrak{p}/1+\mathfrak{p}^2 \subset \kappa_{K_{\mathfrak{p}}} \\ & \Rightarrow 1+\mathfrak{p} =(\text{subgroup of $\kappa_{K_{\mathfrak{p}}}$}) \oplus (1+\mathfrak{p}^2) \end{align}
EDIT: As clarified by the OP, the intended question is the decomposition of $U_{K, \mathfrak p} = \{x \in \mathcal O_K^\times: x \equiv 1 \pmod{\mathfrak p}\}$ instead of $1 + \mathfrak p$.
So we define, for an ideal $I$ of $\mathcal O_K$, the unit group $U_{K, I} \colon = \{x \in \mathcal O_K^\times: x \equiv 1 \pmod I\}$. The question asks whether $U_{K, \mathfrak p}$ is equal to $\mu_p \oplus U_{K, \mathfrak p^2}$.
This is true. Proof: The quotient $U_{K_{\mathfrak p}, \mathfrak q}/U_{K_{\mathfrak p}, \mathfrak q^2}$ (what the OP writes as $1 + \mathfrak q / 1 + \mathfrak q^2$) is isomorphic to $\Bbb Z / p\Bbb Z$ as the extension $K / \Bbb Q$ is totally ramified at $p$. Moreover the injective map $U_{K, \mathfrak p}/U_{K, \mathfrak p^2} \to U_{K_{\mathfrak p}, \mathfrak q}/U_{K_{\mathfrak p}, \mathfrak q^2}$ is not constantly $1$ because the image of a generator of $\mu_p$ is not $1$. Therefore it must be an isomorphism.