Problem 6.1.14 - If $g\in L^{\infty}$, the operator $T$ defined by $Tf = fg$ is bounded on $L^p$ for $1\leq p \leq \infty$. Its operator norm is at most $\|g\|_{\infty}$, with equality if $\mu$ is semifinite.
Attempted proof - Follows the advice from below. For $1\leq p \leq \infty$, $$\|Tf\|_{p}^{p} = \int |Tf|^{p} d\mu = \int |fg|^p d\mu \leq \|g\|_{\infty}^p\|f\|_{p}^{p}$$ by Theorem 6.8. Now, if we take the $p$-th root of this expression we have $$\|Tf\|_{p} \leq \|g\|_{\infty}\|f\|_{p}$$
Not sure if this correct for the first part, any suggestions on this or the remaining parts to prove is greatly appreciated.
For $1\leq p<\infty$, $$\Vert Tf\Vert_p^p =\int |Tf|^p d\mu =\int |fg|^pd\mu\leq \Vert g\Vert_{\infty}^p \Vert f\Vert_p^p,$$ where we have used the fact that $|g| \leq \Vert g\Vert_{\infty}$ $\mu$-a.e. Take $p$th roots to get $$\Vert Tf\Vert_p \leq \Vert g\Vert_{\infty}\Vert f\Vert_p.$$ This is exactly what you wanted for $p$ finite.
For $p=\infty$, first show that $\Vert fg\Vert_{\infty}\leq \Vert g\Vert_{\infty}\Vert f\Vert_{\infty}$, so then $\Vert Tf\Vert_{\infty}\leq \Vert g\Vert_{\infty}\Vert f\Vert_{\infty}.$ Again, this is what you want.
Edit:
An operator $T$ on a normed vector space $(X, \Vert \cdot \Vert)$ is bounded if there is $C>0$ such that $$\Vert Tx \Vert \leq C \Vert x\Vert$$ for all $x \in X$. Equivalently, $$\sup_{\Vert x\Vert \leq 1} \Vert Tx\Vert <\infty.$$
In this case, the norm is $L^p$ norm, and the problem asks you to show that $C$ can be chosen to be $\Vert g\Vert_{\infty}$.