The total number of monohalogenated organic products in the following (including stereoisomers) reaction is
A (Simplest optically active alkene) $$\xrightarrow[(ii) X_2/\Delta]{(i) H_2/Ni/\Delta}$$ 

First, we have to recognise what “the simplest optically active alkene” is. In hydrocarbons the first stereogenic (chiral) carbon atom can appear only when the molecule contains at least five carbon atoms, because with four carbon atoms or fewer some two of the four substituents on any sp3 carbon are necessarily identical. The smallest alkene fulfilling this requirement is $$CH_{2}=CH-CH(CH_{3})-CH_{2}-CH_{3}$$ whose IUPAC name is $$3\text{-methyl-1-pentene}.$$ The italicised carbon in the middle bears four different groups - $$CH_{3},\;H,\;CH_{2}=CH-$$ and $$-CH_{2}CH_{3}$$ - so the alkene is optically active.

Now we pass this alkene through step (i): catalytic hydrogenation with $$H_{2}/Ni$$ at high temperature. The general reaction is stated first:

$$R-CH=CH-R' + H_{2} \;\xrightarrow{Ni,\;\Delta\; R-CH2-CH2-R'}$$

Applying it to our substrate, the double bond between the first two carbons is reduced, giving

$$CH_{3}-CH_{2}-CH(CH_{3})-CH_{2}-CH_{3},$$

which is $$3\text{-methylpentane}.$$ After saturation, the central carbon is bonded to two identical ethyl groups, a methyl group and a hydrogen atom, so the molecule now has an internal plane of symmetry and becomes achiral.

Step (ii) is halogenation with a halogen $$X_{2}$$ in the presence of heat: $$R-H + X_{2} \;\xrightarrow{\Delta\; R-X + HX}.$$ Under these free-radical conditions any hydrogen atom of the alkane may be replaced by one halogen atom, and we have to count every distinct monohalogenated product, including enantiomers when they appear.

Let us label the carbon atoms of $$3\text{-methylpentane}$$ for clarity:

$$C_{1}H_{3} - C_{2}H_{2} - C_{3}H(CH_{3}) - C_{4}H_{2} - C_{5}H_{3}$$
and the branched $$CH_{3}$$ on C3 is $$C_{3}'.$$

Because the molecule is symmetric about C3, the pairs (C1, C5) and (C2, C4) are equivalent. Thus there are only four kinds of hydrogen positions:

Type I : the six terminal hydrogens on C1 or C5.
Type II : the four hydrogens on C2 or C4.
Type III : the single hydrogen on the central carbon C3.
Type IV : the three hydrogens on the side-chain methyl C3′.

We now replace one hydrogen of each type by X and inspect the stereochemistry.

Type I substitution gives $$XCH_{2}-CH_{2}-CH(CH_{3})-CH_{2}-CH_{3}$$ (1-halo-3-methylpentane). Introducing X on one end breaks the internal symmetry, so the formerly achiral carbon C3 now has four different groups:
$$-CH_{2}XCH_{2}, \; -CH_{2}CH_{3}, \; -CH_{3},\; H.$$
C3 therefore becomes chiral, and a pair of enantiomers is produced. Because the two ends are equivalent, the product from C1 and the product from C5 are the same pair. Number of distinct products here = 2.

Type II substitution gives $$CH_{3}-CHX-CH(CH_{3})-CH_{2}-CH_{3}$$ (2-halo-3-methylpentane). Now two carbons are stereogenic: C2 (bearing X) and C3. With two chiral centres and no internal symmetry four stereoisomers are possible: RR, SS, RS and SR. (RR and SS are enantiomeric; RS and SR are another enantiomeric pair.) Substituting at C4 generates exactly the same four structures, so the total number of new products here is 4.

Type III substitution replaces the lone hydrogen on C3 itself: $$CH_{3}-CH_{2}-C(X)(CH_{3})-CH_{2}-CH_{3}.$$
C3 now carries two identical ethyl groups, X and CH3, so it is not chiral; the molecule has no other stereogenic centre. Hence only 1 product arises from this site.

Type IV substitution gives $$CH_{3}-CH_{2}-CH(CH_{2}X)-CH_{2}-CH_{3}.$$ C3 again has two identical ethyl groups and is therefore achiral, so this also contributes just 1 product.

Adding all the distinct mono-halogenated structures together:

$$2\;(\text{from Type I}) + 4\;(\text{from Type II}) + 1\;(\text{from Type III}) + 1\;(\text{from Type IV}) = 8.$$

So, the answer is $$8$$.