Let us show that any line bundle has a smooth metric. There is a general theory, due to Raynaud, that shows how to define formal models from rigid analytic objects. In the present case, $X$ being projective, we may assume that $L$ is ample and consider a closed embedding of $\mathrm{X}$ in a projective space $\mathrm{P}^{n}_{K}$ given by some power $L^{e}$. Let $\mathfrak{X}$ be the Zariski closure of $\mathrm{X}$ in $\mathrm{P}^{n}_{K^{\circ}}$ ; in concrete terms, if $I\subset K[X_{0},\dots,X_{n}]$ is the homogeneous ideal of $i(X)$, $I\cap K^{\circ}[X_{0},\dots,X_{n}]$ is the homogeneous ideal of $\mathfrak{X}$. Let then $\mathfrak{L}$ be the restriction to $\mathfrak{X}$ of the line bundle $\mathscr{O}(1)$. The triple $(\mathfrak{X},\mathfrak{L},e)$ is a model of $L$ and induces a smooth metric on $L$.

Different models can give rise to the same metric. If $\varphi\colon\mathfrak{X}^{\prime}\rightarrow\mathfrak{X}$ is a morphism of models, and $\mathfrak{L}^{\prime}=\varphi^{*}\mathfrak{L}$, then $(\mathfrak{X}^{\prime},\mathfrak{L}^{\prime},e)$ defines the same smooth metric on $L$. Moreover, if two models $(\mathfrak{X}_{i},\mathfrak{L}_{i},e_{i})$, for $i\in\{1,2\}$, define the same metric, there exists a third model $(\mathfrak{X},\mathfrak{L})$, with two morphisms $\varphi_{i}\colon\mathfrak{X}\rightarrow\mathfrak{X}_{i}$ such that the pull-backs $\varphi_{i}^{*}\mathfrak{L}_{i}^{e_{1}e_{2}/e_{i}}$ coincide with $\mathfrak{L}$. More precisely, if two models $\mathfrak{L}$ and $\mathfrak{L}^{\prime}$ of some power $L^{e}$ on a normal model $\mathfrak{X}$ define the same metric, then they are isomorphic. (See, e.g., Lemma 2.2 of [19] ; this may be false for non-normal models ; it suffices that $\mathfrak{X}$ be integrally closed in its generic fiber.)

As a consequence, the set $\overline{\operatorname{Pic}}_{\text{sm}}(\mathrm{X})$ of smooth metrized line bundles is a subgroup of the group $\overline{\operatorname{Pic}}(\mathrm{X})$. The group $\overline{\operatorname{Pic}}_{\text{sm}}(\mathrm{X})$ fits within an exact sequence

$$0\rightarrow\mathscr{C}^{\infty}(\mathrm{X})\rightarrow\overline{\operatorname{Pic}}_{\text{sm}}(\mathrm{X})\rightarrow\operatorname{Pic}(\mathrm{X})\rightarrow 0,$$

the last map is surjective because every line bundle admits a model. If $f\colon\mathrm{Y}\rightarrow\mathrm{X}$ is a morphism, then $f^{*}(\overline{\operatorname{Pic}}_{\text{sm}}(\mathrm{X}))\subset\overline{\operatorname{Pic}}_{\text{sm}}(\mathrm{Y})$.