A new electron enters the orbital when:

The correct option is A $(n+l)$ is minimum
A new electron enters the orbital when $(n+l)$ is minimum.

For example, the 4s sub-energy level is at lower energy than the 3d sub-energy level.

For 3d, $n=3$ and $l=2$. Hence, $(n+l)=3+2=5$

For 4s, $n=4$ and $l=0$. Hence, $(n+l)=4+0=4$

As the value of $(n+l)$ for 4s orbital is less than that of 3d orbital, 4s orbital is filled before 3d orbital.

Hence, option A is correct.