Depends on which unit we choose. For instance, let $K = \mathbb{Q}$ and let $\mathfrak{p}$ be the prime ideal $(2)$ above the rational prime $p=2$.

If we choose the unit $\varepsilon = 1$, then $K(\sqrt{\varepsilon}) = \mathbb{Q}$ and $\mathfrak{p}$ remains unramified.

If we choose the unit $\varepsilon = -1$, then $K(\sqrt{\varepsilon}) = \mathbb{Q}(i)$ and $\mathfrak{p}$ is the square of the ideal $(1-i)$, since $2 = i(1-i)^2$.