SN2 Reaction:
SN2 (substitution nucleophilic bimolecular) reaction is a type of nucleophilic substitution reaction in which a nucleophile displaces a leaving group from a carbon atom, resulting in the substitution of one functional group by another. In SN2 reactions, the nucleophile attacks the carbon center simultaneously as the leaving group departs.

Reactivity in SN2 Reaction:
The reactivity of alkyl halides in SN2 reactions depends on a few factors, including the nature of the leaving group, the steric hindrance around the carbon atom, and the strength of the nucleophile.

Analysis of the Statements:
a) 1-iodopropane reacts more rapidly than 2-iodo propane in SN2 reaction:
- This statement is correct. In SN2 reactions, the reactivity of alkyl halides increases with the increase in the size of the halogen atom. Iodide is larger and more polarizable than bromide, which makes it a better leaving group. Therefore, 1-iodopropane reacts more rapidly than 2-iodopropane in SN2 reactions.

b) 1-chloropropane has greater reactivity than 1-bromopropane in SN2 reaction:
- This statement is incorrect. In SN2 reactions, the reactivity of alkyl halides increases with the increase in the size of the halogen atom. Bromide is larger and more polarizable than chloride, which makes it a better leaving group. Therefore, 1-bromopropane has greater reactivity than 1-chloropropane in SN2 reactions.

c) 1-bromopropane reacts faster than (CH3)3CCH2Br under SN2 condition:
- This statement is correct. In SN2 reactions, the reactivity of alkyl halides decreases with an increase in steric hindrance around the carbon atom. The (CH3)3CCH2Br has a tert-butyl group attached to the carbon atom, which creates significant steric hindrance. Therefore, 1-bromopropane reacts faster than (CH3)3CCH2Br under SN2 conditions.

d) PCH3OC6H4Br has greater reactivity than p-O2NC6H4CH2Br in SN2 reaction:
- This statement is incorrect. In SN2 reactions, the reactivity of alkyl halides decreases with an increase in steric hindrance around the carbon atom. The p-O2NC6H4CH2Br has a larger substituent attached to the carbon atom, which creates more steric hindrance. Therefore, PCH3OC6H4Br has greater reactivity than p-O2NC6H4CH2Br in SN2 reactions.

Conclusion:
Based on the analysis, the correct statements are a) 1-iodopropane reacts more rapidly than 2-iodo propane in SN2 reaction and c) 1-bromopropane reacts faster than (CH3)3CCH2Br under SN2 conditions.