Step 1 - Classify the $-OH$ group.
The $-OH$ group of phenol is a strongly ACTIVATING, ORTHO/PARA-DIRECTING group.
The lone pair on oxygen is donated into the ring by resonance, so electron density is
highest at the ortho and para positions.

Step 2 - Note the reaction conditions.
DILUTE $HNO_3$ and LOW temperature (about 298 K) are mild conditions. They can bring
about only MONO-nitration.

Step 3 - Write the products.
$$C_6H_5OH \xrightarrow[\text{298 K}]{\text{dil. } HNO_3} \text{o-nitrophenol} + \text{p-nitrophenol}$$
A mixture of ortho- and para-nitrophenol is obtained (roughly in equal amounts).
These two are separated by steam distillation, since o-nitrophenol has intramolecular
hydrogen bonding (chelation) and is steam volatile, while p-nitrophenol has
intermolecular hydrogen bonding and is not.

Step 4 - Reject option (A).
2,4,6-Trinitrophenol (picric acid) requires CONCENTRATED $HNO_3$ (with $H_2SO_4$),
not dilute acid at low temperature.

Answer: (D) ortho-and para-nitrophenol