The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is Option D. Let's break down why this is the case by understanding both the $$S_N 1$$ and $$S_N 2$$ reactions and their effects on the chirality of the alkyl halide.

$$S_N 2$$ Reaction:

The $$S_N 2$$ (Substitution Nucleophilic Bimolecular) reaction is a one-step process where the bond formation (nucleophile attacking the substrate) and the bond-breaking (leaving group leaving the substrate) processes occur simultaneously. This reaction type is associated with an inversion of configuration at the chiral center. This means that if the reactant molecule is chiral and the nucleophile attacks from the side opposite to the leaving group, the product will have the opposite configuration to the reactant. This phenomenon is known as Walden inversion. Therefore, it does not lead to racemisation but rather to inversion of the stereochemistry at the chiral center.

$$S_N 1$$ Reaction:

The $$S_N 1$$ (Substitution Nucleophilic Unimolecular) reaction is a two-step process. The first step involves the formation of a carbocation intermediate by the departure of the leaving group. This intermediate is planar, and hence, the nucleophile can attack from either side of the plane with equal probability. This results in the formation of products with both configurations - one retaining the original configuration and the other being its mirror image (inverted configuration). As both enantiomers are formed, the process leads to racemisation, especially in situations where the carbocation can freely rotate and is attacked with equal probability from both sides. Over time, if this reaction goes to completion and there's no bias in nucleophile attack direction, you'd expect a racemic mixture of products.

Thus, Option D is correct: Racemisation occurs in $$S_N 1$$ reaction, and inversion occurs in the $$S_N 2$$ reaction.