We start with a perfect number (which is equal to the sum of its divisors, except itself) different from $6$ and calculate its sum of digits. Then, we calculate the sum of digits of the new number and so on. Prove that we will eventually get $1$.
For even perfect numbers this equation is easy since we know even perfect numbers are of the form $2^{n-1}(2^n - 1)$ where $2^n - 1$ is prime. We can then check that $2^{n-1}(2^n - 1)\equiv 1\mod 9$ and since the digit sum of a number is congruent to the number itself mod 9, the result follows.
However, I'm not sure how to do this for odd perfect numbers or if this is even possible. It's possible that the question above has a mistake, but I wanted to double check.
Since there are no currently known odd perfect numbers, and whether or not they even exist is still an open question, I suspect your question is referring to only even perfect numbers. It's also likely not referring to odd perfect numbers because there are certain known congruence properties of any that might be found which don't support what it's asking.
In particular, according to Wikipedia's Odd perfect numbers article section, the third bullet point of what any such number $N$ must satisfy states that
Having $N \equiv 1 \pmod{9}$ is possible with the first congruence if, since $\operatorname{lcm}(12, 9) = 36$, we have $N \equiv 1 \pmod{36}$. However, the second and third congruences have the equivalence values & the moduli both being multiples of $9$, so then $N \equiv 0 \pmod{9}$.
Note a related post here is Proof that an odd perfect number must be in the form $ 36k+9$ or $12k+1$.