Answer
Option (b) is the correct answer.
Work Step by Step
When the elevator is moving with a constant velocity, the scale shows the person's true weight. So $W_{true}=mg=600N$
As the elevator decelerates $(a\lt 0)$ while still going upward (in the direction of normal force $F_N$), according to Newton's 2nd law, we have $$F_N-mg=-ma$$
As the right side is negative, we have $F_N\lt mg$, so $F_N\lt600N$
When the elevator accelerates $(a\gt 0)$ on the way down (in the direction of the person's weight), according to Newton's 2nd law, we have $$mg-F_N=ma$$
As the right side is positive, we have $F_N\lt mg$, so $F_N\lt600N$
Therefore, option (b) is the correct answer.
