An atom has electronic configurations 2, 8, and 7. 

To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.) 

In modern periodic table, the elements in similar group show similar properties. The given element has electronic configuration 2, 8, 7. Since the last shell contains 7 electrons it will be in group 17.

Now consider the options;

Ar- 2, 8, 8

F- 2, 7

P- 2, 8, 5

N- 2, 5

The electronic configuration of chlorine is $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$.
The valence electrons in chlorine are 7.
We have to find the chemically similar element to chlorine.
The given elements are Nitrogen (7), Fluorine (9), Phosphorous (15), Argon (18).
To know about the valence electrons present in Nitrogen (7), Fluorine (9), Phosphorous (15), and Argon (18). We should know the electronic configuration of those elements.
Electronic configuration of Nitrogen $\left(7\right)=1s^{2}2s^{2}2p^3$
Electronic configuration of Fluorine $\left(9\right)=1s^{2}2s^{2}2p^5$
Electronic configuration of Phosphorus (15) $=1s^{2}2s^{2}2p^{6}3s^{2}3p^3$
Electronic configuration of Argon (18) $=1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

As a result, the valence electrons for the elements nitrogen, phosphorus, and argon are 5, 5, and 8 correspondingly. Fluorine, however, has seven valence electrons.
Both fluorine and chlorine have the same amount of valence electrons and a comparable electrical structure.

In these only fluorine with atomic number 9 goes to the same group, 7. Therefore fluorine is the similar element.

Hence, option B is correct.