I will refer to two events as $A$ and $B$ and refer to person $B$ as person $1$ and $A$ as person $2$.
If you are applying Bayes' theorem to find posterior probability of an event, first check what the event is and what the given condition is. The given condition in this case is that person $2$ had to roll twice to get a number equal to or more than person $1$. Say that event is $B$. If the event that person $1$ rolls four is $A$ then you are trying to find the probability $P(A\mid B)$. You need to first find $P(A \cap B)$ and $P(B)$. Please note that $P(B)$ is unconditional probability of person $2$ having to roll twice to get equal to or more than person $1$.
If person $1$ got one, there is no way person $2$ can roll less than that in the first roll. So there is no second roll required.
If person $1$ rolls two, probability that it takes a second roll for person $2$ to roll two or more,
$ \displaystyle \frac{1}{6} \cdot \frac{5}{6}$
Similarly if person $1$ rolls $3, ~$ it is
$ \displaystyle~ \frac{2}{6} \cdot \frac{4}{6}$
So if you add all cases and multiply by the probability of person $1$ rolling each number,
$ \displaystyle P(B) = \frac{1 \cdot 5 + 2 \cdot 4 + 3 \cdot 3 + 4 \cdot 2 + 5 \cdot 1}{6 \cdot 6 \cdot 6}$
Now what is $P(A \cap B)$? That is person $1$ rolls four and it takes a second roll for person $2$ to roll four or more.
Then finally, you can apply the formula $~P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}$