I am trying to show \begin{align*} \mathbb{E}(\sup_{t \in [0,T]} \exp(B_t)) \leq 4 \mathbb{E} \exp(B_T) \end{align*} For T>0.
Since there is a $4$, I guessed that it might be an application of the Doob $L^p$ inequality for $p=2$ because $||M_n^{\ast}||^2_2 \leq \left(\frac{2}{2-1}\right)^2 ||M_n||^2_2$ looks exactly like what we need. I am having trouble seeing how this might be an application of Doob, and maybe I need another tool.
The process $M_t := \exp(B_t)$ is a submartingale. Indeed, for all $s \leq t$,
\begin{align*} \mathbb{E}(\exp(B_t) \mid \mathcal{F}_s) &= \exp(B_s) \mathbb{E}(\exp(B_t-B_s) \mid \mathcal{F}_s) \\ &= \exp(B_s) \mathbb{E}(\exp(B_t-B_s)) \\ &= \exp(B_s) \exp(t-s) \\ &\geq \exp(B_s) \end{align*}
where we used the independence of the increments and the fact that $B_t-B_s \sim N(0,t-s)$. Moreover, the process $(M_t)_{t \geq 0}$ is clearly non-negative and has continuous sample paths (with probability $1$). Since Doob's maximal inequality applies to non-negative submartingales with continuous sample paths, this means that we may apply Doob's maximal inequality to derive the desired inequality.
Remark: Here is an alternative approach (without Doob's maximal inequality): Because of the monotonicity of $\exp$, we have
$$\sup_{t \in [0,T]} \exp(B_t) = \exp \left( \sup_{t \in [0,T]} B_t \right).$$
The distribution of $\sup_{t \in [0,T]} B_t$ is known from the so-called reflection principle: $$\sup_{t \in [0,T]} B_t \sim |B_T|.$$ Hence, $$\mathbb{E}\left( \sup_{t \in [0,T]} \exp(B_t) \right) = \mathbb{E}\exp(|B_T|).$$ From $$\exp(|x|) \leq \exp(x) + \exp(-x)$$ and the fact that $B_T$ has the same distribution as $-B_T$, we conclude that $$\mathbb{E}\left( \sup_{t \in [0,T]} \exp(B_t) \right) \leq \mathbb{E}\exp(B_T)+\mathbb{E}\exp(-B_T) = 2 \mathbb{E}\exp(B_T).$$