What would be the compositum $K \cdot \mathbb{Q}_p(\sqrt u)$? Consider $K$ be an unramified extension of $p$-adic field $\mathbb{Q}_p$ of degree $n$.

I want to compute the compositum $K \cdot \mathbb{Q}_p(\sqrt u)$, where $u^2=-1$.

Since $K$ is unramified extension, we have $K \subseteq \mathbb{Q}_p(\zeta_n)$.
So we have $K \cdot \mathbb{Q}_p(\sqrt u) \subseteq \mathbb{Q}_p(\zeta_n) \cdot \mathbb{Q}_p(\sqrt u).$
What would be the exact computation ?

Is $K \cdot \mathbb{Q}_p(\sqrt u)=K(\sqrt u)$ ?

 A: As reuns points out in a comment, if $L$ is any field, and $K \vert L$ an extension, and both $K$ and an element $\alpha$ are contained in a common extension $F \vert L$ (say, e.g., $F = $ an algebraic closure of $L$ or $K$), then the compositum is
$$K.L(\alpha) = K(\alpha).$$
This equality is so immediate from the definitions that I would have a hard time explaining it further.
In your case $L=\mathbb Q_p$ and $\alpha$ is some primitive root of unity; I am still not sure if you mean a fourth root of unity, i.e. $\alpha^2=-1$, i.e. $\alpha=\zeta_4$, or an eighth root of unity, i.e. $\alpha^4=-1$, i.e. $\alpha=\zeta_8$, but in both cases it is straightforward to determine that field.
Case 1: $p$ odd. Then because $2^n$-th roots of unity generate unramified extensions, it's immediate that $$\zeta_4 \in K \Leftrightarrow 4 \vert (p^n-1) \Leftrightarrow p \equiv 1 (4) \text{ or } n \text{ is even}$$
$$\zeta_8 \in K \Leftrightarrow 8 \vert (p^n-1)\Leftrightarrow p \equiv 1 (8) \text{ or } n \text{ is even}$$
meaning that precisely in these cases, the compositum $=K(\alpha)$ is just $K$ itself. This also tells us (because for each degree $d$ there is a unique unramified extension of $\mathbb Q_p$ in a given algebraic closure, and they are all cyclic) that in all remaining cases, $K(\alpha) \vert K$ is of degree $2$ i.e. the unique quadratic unramified extension of $K$.
Case 2: $p=2$. Now $\mathbb Q_2(\zeta_4)$ and $\mathbb Q_2(\zeta_8)$ are both totally ramified extensions, of degree $2$ and $4$ respectively. But since $K\vert \mathbb Q_2$ is unramified, and regardless of what else it is, ramification theory forces $K(\zeta_4) \vert K$ and $K(\zeta_8) \vert K$ to be totally ramified extensions of degree $2$ and $4$, respectively, as well.
