Let's just write $\|f\|$ for the infinity norm. We have that $\|f\|$ is defined as the essential sup of the values that $f$ takes. But then $|f(x)|^p\le\|f\| ^p$ for any $x\in X$ except in a set of measure 0.
Hence, what you have is $\int|f|^p\le \int\|f\| ^p$, because the set where the inequality between $|f|$ and $\|f\|$ is reversed is negligible. And, since $\|f\| ^p$ is a constant, we can move it out of the integral.
(Note $\|f\|$ could very well be infinite, but in that case the inequality you need is obvious, because anything is $\le\infty$. So, only the case where $\|f\| <\infty$ needs arguing.)
Ok, I was using the wrong norm. You have $\|f\|$ defined as the infimum of $\|h\|_ {\rm sup}$ where $h$ ranges over all (measurable) functions that agree with $f$ a.e.
Fix any such $h$. Then $|f|=|h|$ a.e., so $|f|^p=|h|^p\le \|h\|^ p_ {\rm sup}$ a.e. So $\int |f|^p=\int |h|^p\le \|h\|^p_ {\rm sup}\int 1$, as above.
Now take the infimum of the right hand side.