Let $(M_n)$ a martingale s.t. $p^{th}$ moment is uniformly bounded. Then $(M_n)$ is uniformly integrable. (So, I'm not sure if it's for all $p^{th}$ moment or at least one).
So let $p$ s.t. $$\mathbb E[|M_n|^p]\leq C.$$ Then, $$\mathbb E[|M_n|]\leq C.$$ Now, $$\sup_{n\in\mathbb N}\mathbb E[|M_n|\boldsymbol 1_{|M_n|\geq K}]\leq C.$$
Why this would implies that $$\lim_{P\to \infty }\sup_{n\in\mathbb N}\mathbb E[|M_n|\boldsymbol 1_{|M_n|\geq K}]=0\ \ ?$$
We require $p>1$, the statement does not hold for $p=1$. In particular, by trying to prove this by only using that $\mathbb E[|M_n|]$ is bounded, you are doomed to fail. However, we can try directly by Holder's inequality:
$$ \mathbb E[|M_n|\mathbf 1_{|M_n| \ge K}] \le \left(\mathbb E[|M_n|^p]\right)^{1/p}\left(\mathbb E[\mathbf 1_{|M_n| \ge K}^q]\right)^{1/p} \le C^{1/q}\mathbb P(|M_n|\ge K)^{1/q},$$
where $q$ satisfies $\frac1p + \frac1q = 1$. Applying Chebyshev's inequality, we have
$$\mathbb P(|M_n| \ge K) \le \frac{\mathbb E[|M_n|^p]}{K^p} \le \frac{C}{K^p},$$
and plugging this back into our first inequality, we find
$$ \mathbb E[|M_n|\mathbf 1_{|M_n| \ge K}] \le C^{1/p}\left(\frac{C}{K^p}\right)^{1/q} = \frac{C}{K^{p/q}}, $$ which is independent of $n$ and tends to zero as $K\to\infty$.