= Solution
Put $V_t=\langle X\rangle_t$ and $S_t=\sup_{s\leq t}|X_s|$. Suppose first that $|X|\leq K$. The <Itô formula> for $|x|^p$, which is twice continuously differentiable for $p\geq2$, gives
$$
|X_t|^p=p\int_0^t|X_s|^{p-2}X_s\,dX_s+\frac{p(p-1)}2\int_0^t|X_s|^{p-2}\,dV_s.
$$
For $p=2$, the second derivative is interpreted as the constant $2$. The stochastic term has mean zero: its integrand is bounded, and $\mathbb EV_t<\infty$ makes it a square-integrable <martingale>. Consequently
$$
\mathbb E|X_t|^p\leq\frac{p(p-1)}2\mathbb E(S_t^{p-2}V_t).
$$
This is the first required estimate.
A bounded <local martingale> is a true <martingale>. Apply the allowed <Doob Lp maximal inequality>, and then <Hölder's inequality> with conjugate exponents $p/(p-2)$ and $p/2$ for $p>2$. With $D_p=(p/(p-1))^p$, this yields
$$
\mathbb ES_t^p\leq D_p\frac{p(p-1)}2(\mathbb ES_t^p)^{(p-2)/p}(\mathbb EV_t^{p/2})^{2/p}.
$$
If the left side is zero there is nothing to prove; otherwise divide by its indicated power and raise to $p/2$. For $p=2$, the same conclusion follows directly from $\mathbb EX_t^2=\mathbb EV_t$ and Doob's inequality. A usable constant is therefore
$$
\boxed{\mathbb ES_t^p\leq C_p\mathbb EV_t^{p/2},\qquad C_p=\left[\frac{p(p-1)}2\left(\frac p{p-1}\right)^p\right]^{p/2}.}
$$
In particular $C_2=4$. This is the <upper maximal moment bound for a continuous local martingale>.
For an unbounded <continuous local martingale>, stop at $\tau_n=\inf\{s:|X_s|\geq n\}$. Its stopped path is bounded, and its bracket at $t$ is $V_{t\wedge\tau_n}\leq V_t$. The proved inequality gives a bound by $C_p\mathbb EV_t^{p/2}$, independent of $n$. Continuity makes $\tau_n\uparrow\infty$, and the stopped maxima increase to $S_t$. <Monotone convergence> proves \b[the same inequality for the original unbounded process], with the same constant. This localization step also shows that the assumed bracket moments supply all the needed maximal moments.
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