Take $p=3$ and $C\subset \mathbb{P}^2$ with equation $y^2 z=x^3 - t z^3$ where $t\in K$ is not a cube. Then $C$ is regular but, putting $L:=K(t^{1/3})$, $C_L$ is isomorphic to the usual cuspidal cubic (explicitly, the equation becomes $y^2 z=(x - t^{1/3} z)^3$).

Thus, putting $J:=\mathrm{Pic}^0_{C/K}$, it follows that $J_L=\mathrm{Pic}^0_{C_L/L}$ is isomorphic to $\mathbb{G}_{a,L}$. In other words, $J$ is a form of $\mathbb{G}_{a,K}$, in particular smooth, one-dimensional and killed by $p$.

The natural map $\mathrm{Pic}^0(C)\to J(K)$ is injective, so $\mathrm{Pic}^0(C)=\mathrm{Pic}^0(C)[3]$. It is even bijective since $C$ has a rational point, namely $(0:1:0)$, so we get a counterexample if $J(K)$ is infinite, which is certainly the case if $K$ is large (aka fertile or ample). So, explicitly, we can take $K=\mathbb{F}_3(\!(u)\!)$ and $t=-u^2$.

Take $p=3$ and $C\subset \mathbb{P}^2$ with equation $y^2 z=x^3 - t z^3$ where $t\in K$ is not a cube. Then $C$ is regular but, putting $L:=K(t^{1/3})$, $C_L$ is isomorphic to the usual cuspidal cubic (explicitly, the equation becomes $y^2 z=(x - t^{1/3} z)^3$).

Thus, putting $J:=\mathrm{Pic}^0_{C/K}$, it follows that $J_L=\mathrm{Pic}^0_{C_L/L}$ is isomorphic to $\mathbb{G}_{a,L}$. In other words, $J$ is a form of $\mathbb{G}_{a,K}$, in particular smooth, one-dimensional and killed by $p$.

The natural map $\mathrm{Pic}^0(C)\to J(K)$ is injective, so $\mathrm{Pic}^0(C)=\mathrm{Pic}^0(C)[3]$. It is even bijective since $C$ has a rational point, namely $(0:1:0)$, so we get a counterexample if $J(K)$ is infinite, which is certainly the case if $K$ is large (aka fertile or ample). So, explicitly, we can take for instance $K=\mathbb{F}_3(\!(t)\!)$.

Take $p=3$ and $C\subset \mathbb{P}^2$ with equation $y^2 z=x^3 - t z^3$ where $t\in K$ is not a cube. Then $C$ is regular but, over $K(t^{1/3})$, $C$ becomes isomorphic to the usual cuspidal cubic (explicitly, the equation becomes $y^2 z=(x - t^{1/3} z)^3$) whose $\mathrm{Pic}^0$ is isomorphic to $\mathbb{G}_a$, in particular killed by $3$, and not finite.

Take $p=3$ and $C\subset \mathbb{P}^2$ with equation $y^2 z=x^3 - t z^3$ where $t\in K$ is not a cube. Then $C$ is regular but, putting $L:=K(t^{1/3})$, $C_L$ is isomorphic to the usual cuspidal cubic (explicitly, the equation becomes $y^2 z=(x - t^{1/3} z)^3$).

Thus, putting $J:=\mathrm{Pic}^0_{C/K}$, it follows that $J_L=\mathrm{Pic}^0_{C_L/L}$ is isomorphic to $\mathbb{G}_{a,L}$. In other words, $J$ is a form of $\mathbb{G}_{a,K}$, in particular smooth, one-dimensional and killed by $p$.

The natural map $\mathrm{Pic}^0(C)\to J(K)$ is injective, so $\mathrm{Pic}^0(C)=\mathrm{Pic}^0(C)[3]$. It is even bijective since $C$ has a rational point, namely $(0:1:0)$, so we get a counterexample if $J(K)$ is infinite, which is certainly the case if $K$ is large (aka fertile or ample). So, explicitly, we can take $K=\mathbb{F}_3(\!(u)\!)$ and $t=-u^2$.