Step 1: Approach
Label every orbital with its two quantum numbers and make a quick table of $n+l$.
Step 2: The table
- 3s has $n=3, l=0$, sum $3$.
- 3p has $n=3, l=1$, sum $4$.
- 2p has $n=2, l=1$, sum $3$.
- 4s has $n=4, l=0$, sum $4$.
- 3d has $n=3, l=2$, sum $5$.
- 4p has $n=4, l=1$, sum $5$.
Step 3: Match up the sums
Pair (A) gives 3 and 4, pair (B) gives 3 and 4, pair (D) gives 3 and 4. Only pair (C) gives 5 and 5. When the sums are equal, the orbital with the smaller $n$ is filled first, so 3d is lower than 4p, but the (n + l) values are the same.
Final Answer:
3d and 4p both have $n+l=5$, so the answer is option (C).
\[ \boxed{\text{3d and 4p}} \]