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.