How many electrons with l = 1 does Si in its ground state have?

I was solving practice problems for electron configuration and periodic table, and I got stuck through a question:

An atom of silicon in its ground state has how many electrons with quantum number $l = 1$?
a) 14
b) 2
c) 8
d) 6
c) 28

According to the solution the answer is c: 8 electrons, but that wasn't my answer.

My analysis:

1. Silicon is a chemical element with symbol $\ce{Si}$ and atomic number 14, this is its electron configuration:

$$\mathrm{1s^22s^22p^63s^23p^2}$$

1. The quantum number $l = 1$ corresponds to $\mathrm{p}$ levels only. So we have:

$$\mathrm{2p^63p^2}$$

1. The total number of orbitals is: $6 + 2 = 8$ orbitals.

2. Since each orbital has 2 electrons, the total number of electrons is $8 \times 2 = 16$ electrons.

What's wrong with my answer? Maybe the question has a wrong answer?

I was also stuck with questions like these in our exams and I learned through the internet that the superscripts are actually the number of electrons and that $l=1$ refers to p-orbital. In the electronic configuration of Silicon, consider only the p-orbital, since it is aking for electrons that have p-orbitals (l=1)
$2\mathrm{p}^6$ and $3\mathrm{p}^2$ are sub-orbitals of p-orbital. Since 2p have 6 electrons in it and 3p have 2 electrons in it, there are total of 8 electrons that have $l=1$ as one of their quantum numbers.