When a small quantity of naphthalene is added to benzene, the freezing point of benzene decreases. This phenomenon is due to the colligative property known as freezing point depression. The addition of a solute, such as naphthalene, disrupts the orderly arrangement of solvent molecules in the solid phase, thereby lowering the freezing point.

Mathematically, the decrease in freezing point ($\Delta T_f$) can be represented by the equation:

$$ \Delta T_f = i \cdot K_f \cdot m $$

where:

$i$ is the van't Hoff factor (which is 1 for naphthalene as it does not dissociate in solution),

$K_f$ is the cryoscopic constant of the solvent (benzene),

$m$ is the molality of the solution.

Thus, the correct option is:

Option B: Decreases