Let $p$ be prime and $p \mid n$ . Prove that
$$ \phi(n) \le \frac{n(p-1)}{p} $$
We're given $p \mid n$. This means that multiples of $p$, up to the value of $n$, are not coprime with $n$. There are $n/p$ of these multiples.
Since we don't know about any other possible factors of $n$, we can only say the quantity of numbers between 1 and $n$ that are coprime to $n$, which is $\phi(n)$, is less than or equal to $n - n/p$.
And so
$$ \phi(n) \le n - \frac{n}{p} = \frac{np - n}{p} = \frac{n(p-1)}{p} $$
That is
$$ \phi(n) \le \frac{n(p-1)}{p} $$