The IUPAC name of the complex [Pt(NH$$_3$$)$$_2$$Cl(NH$$_2$$CH$$_3$$)]Cl is

We first examine the complete formula $$[\,\text{Pt}(NH_3)_2Cl(NH_2CH_3)\,]Cl$$. The species inside the square brackets is the coordination entity, while the chloride written outside is the counter-ion.

To decide the oxidation state of platinum, we recall the rule:

$$$\text{(Sum of charges of metal and ligands inside the bracket)} = \text{Charge on the complex ion}$$$

We have two neutral $$NH_3$$ ligands, one neutral $$NH_2CH_3$$ ligand, and one $$Cl^-$$ ligand with charge $$-1$$. Let the oxidation number of platinum be $$x$$. The overall charge on the complex ion is $$+1$$, because one external $$Cl^-$$ balances it to give the electrically neutral salt.

Writing the equation,

$$x + 0 + 0 + (-1) = +1$$

So,

$$x - 1 = 1 \; \Longrightarrow \; x = +2$$

Hence platinum is in the $$+2$$ oxidation state, and we shall denote it as platinum(II) in the name.

Now we list the ligands present inside the coordination sphere:

(i) $$NH_3$$  →  “ammine” (there are two such ligands, so the prefix “di-” is used)
(ii) $$Cl^-$$  →  “chlorido” (only one, so no prefix other than “mono-”, which is normally omitted)
(iii) $$NH_2CH_3$$  →  “methanamine” (IUPAC name for methylamine; only one present)

While arranging ligand names, IUPAC rules state that the alphabetical order is decided by the first letter of the ligand name, ignoring any numerical prefixes such as di-, tri-, etc. Therefore the order will be:

ammine (A) < chlorido (C) < methanamine (M)

Consequently, the ligand part of the name is written as

“diamminechlorido(methanamine)”

Finally we attach the name of the metal with its oxidation state in Roman numerals inside parentheses, followed by the name of the counter-ion:

$$$\text{diamminechlorido(methanamine) platinum(II) chloride}$$$

Spacing the words exactly as per IUPAC convention, we obtain

Diamminechlorido (methanamine) platinum (II) chloride

This corresponds precisely to Option A.

Hence, the correct answer is Option A.