The equation which is balanced and represents the correct product(s) is:

We examine every alternative one by one, always checking two things: first, whether the skeleton is stoichiometrically balanced (atoms as well as total charge); second, whether the chemistry that has been written is actually what happens under the stated conditions.

We have for option A

$$Li_2O + 2KCl \;\rightarrow\; 2LiCl + K_2O$$

Atom count:

Left side - $$Li=2,\;O=1,\;K=2,\;Cl=2$$
Right side - $$Li=2,\;O=1,\;K=2,\;Cl=2$$

So the equation is algebraically balanced. However an exchange between two normal, non-volatile salts in the solid state will proceed only if one of the possible products is a gas, a precipitate or a weak electrolyte. Both $$LiCl$$ and $$K_2O$$ are ordinary salts; no driving force such as volatility or precipitation is present, and lattice-energy considerations in fact favour the original combination rather than the exchanged one (the small highly charged $$O^{2-}$$ ion prefers the smaller $$Li^+$$ ion). Hence the reaction does not take place; the equation, although arithmetically balanced, does not represent the correct chemistry.

Now option B:

$$[CoCl(NH_3)_5]^+ + 5H^+ \;\rightarrow\; Co^{2+} + 5NH_4^+ + Cl^-$$

First, we verify the atom balance.

Left - $$Co=1,\;Cl=1,\;N=5,\;H=5\;(\text{outside the ligands}) + 15\;(\text{in } 5NH_3)=20$$

Right - $$Co=1,\;Cl=1,\;N=5,\;H=5\times4(\text{in }NH_4^+)=20$$

Hence atoms match. Now for the total charge:

Left-hand charge = $$\;(+1) + 5(+1)= +6$$

Right-hand charge = $$(+2)+5(+1)+(-1)= +6$$

The charges also match. From the chemical point of view, concentrated acid protonates each $$NH_3$$ ligand to form $$NH_4^+$$; the mutually bound chloride leaves as $$Cl^-$$ and the cobalt centre is reduced from +3 to +2 by internal redox in which the coordinated ammonia is very slowly oxidised (no external reducing agent is required because the process is intramolecular). Thus the products quoted are exactly those observed. The equation is therefore both balanced and chemically correct.

Consider option C:

$$[Mg(H_2O)_6]^{2+} + (EDTA)^{4-} \;\xrightarrow{\text{excess NaOH}}\; [Mg(EDTA)]^{2-} + 6H_2O$$

The arithmetic balance is fine, but the experimental fact is that when a very strong excess of $$NaOH$$ is present, the hexaaquamanganese(II) ion precipitates as $$Mg(OH)_2$$ long before the EDTA can capture it. In other words, under “excess NaOH” conditions the dominant product is a solid hydroxide, not the soluble EDTA chelate. So the equation does not describe the real outcome and must be rejected.

Finally option D:

$$CuSO_4 + 4KCN \;\rightarrow\; K_2[Cu(CN)_4] + K_2SO_4$$

At a first glance the atoms appear balanced (there are four $$K$$ atoms, one $$Cu$$ atom, one $$S$$ atom, four $$CN$$ groups and four $$O$$ atoms on each side). Nevertheless, when excess cyanide reacts with $$Cu^{2+}$$ the sequence is:

$$Cu^{2+} + 2CN^- \;\rightarrow\; Cu(CN)_2\!\downarrow$$ (a white precipitate)

$$Cu(CN)_2 + 2CN^- \;\rightarrow\; [Cu(CN)_4]^{3-} + Cu^+$$

The $$Cu^+$$ liberated then combines with another portion of the complex to give a mixture whose overall composition corresponds to $$K_3[Cu(CN)_4]$$, never $$K_2[Cu(CN)_4]$$. Moreover, the accompanying sulphate remains as $$K_2SO_4$$ only after a second equivalent of $$KCN$$ is introduced. Hence the single, neatly written equation in the option does not match the actual multi-step chemistry; the stoichiometry of potassium ions is also wrong. Thus option D is not acceptable.

Out of the four alternatives, only option B satisfies both the mathematical balance and the true chemical behaviour.

Hence, the correct answer is Option B.