Let $(\Omega, (\mathcal{F}_t), P)$ be a filtered probability space.
Let $1\leq p\leq \infty$ and $M$ be a $p$-integrable $(\mathcal{F}_t)$-martingale closed by $M_\infty$. What do we need from $(\mathcal{F}_t)$, $p$, and $M_\infty$ (e.g. right-continuity of filtration, $p<\infty$, $p$-integrability of $M_\infty$) in order to have $$ \sup_{0\leq t<\infty}\Vert M_t\Vert_p = \Vert M_\infty\Vert_p? $$
Nothing more is needed. Assume for the moment that $M_\infty\in L^p$. $M$ admits a right continuous modification, call it $X$. The process $X$ is a martingale with respect to $(\mathcal F_{t+})$. Being a modification of $M$, $X$ is also closed by $M_\infty$, and because $X$ is right continuous, you have $\lim_{t\to\infty} X_t=M_\infty$ a.s. and in $L^p$. Because $X$ is a modification of $M$: $$ \sup_{0\le t<\infty}\|M_t\|_p=\sup_{0\le t<\infty}\|X_t\|_p. $$ Also, $t\mapsto \|X_t\|_p$ increases because $|X_t|^p$ is a submartingale. It follows that $$ \sup_{0\le t<\infty}\|M_t\|_p=\sup_{0\le t<\infty}\|X_t\|_p=\lim_{t\to\infty}\|X_t\|_p =\|M_\infty\|_p. $$
The 1-integrability of $M_\infty$ is implicit in "$M$ is closed by $M_\infty$". For $p>1$, the $p$-integrability is not needed. Indeed, by Fatou and the previously noted monotonicity, $\|M_\infty\|_p\le\sup_{0\le t<\infty}\|M_t\|_p$. And if this last inequality is strict, then $M_\infty$ is $p$-integrable.