Let $m$ be an even perfect number, that is $m = 2^{n - 1}(2^n - 1)$ where $2^n - 1$ is prime and $n \ge 2$.
Prove that
$$ \phi (m) = 2^{n - 1}(2^{n-1} - 1) $$
We want to show that $2^{n-1}$ and $2^n-1$ are coprime. Since $2^n-1$ is prime, and $n\ge2$, then $2^n-1 \ge 3$ is an odd prime. The only prime factor of $2^{n-1}$ is 2. This means $2^{n-1}$ and $2^n-1$ are coprime.
Having established $2^{n-1}$ and $2^n-1$ are coprime, we can use the multiplicity of $\phi$
$$ \begin{align} \phi(2^{n - 1}(2^n - 1)) & = \phi(2^{n - 1}) \times \phi (2^n - 1) \\ \\ & = 2^{n-2} \times (2^n-2) \\ \\ & = 2^{n-1} \times (2^{n-1}-1) \end{align} $$
This is the desired result.