Among statements (a) - (d), the correct ones are:
(a) Decomposition of hydrogen peroxide gives di-oxygen.
(b) Like hydrogen peroxide, compounds, such as KClO$$_3$$, Pb(NO$$_3$$) and NaNO$$_3$$ when heated liberate di-oxygen.
(c) 2-Ethylanthraquinone is useful for the industrial preparation of hydrogen peroxide.
(d) Hydrogen peroxide is used for the manufacture of sodium perborate.

We have four independent statements (a) to (d). We shall examine each one carefully, deriving or recalling the required chemical equations so that every algebraic-type manipulation is clear.

First, consider statement (a): “Decomposition of hydrogen peroxide gives di-oxygen.” The well-known disproportionation reaction of hydrogen peroxide can be written as

$$2\,\mathrm{H_2O_2}\;\longrightarrow\;2\,\mathrm{H_2O} \;+\; \mathrm{O_2}.$$

Every coefficient is an integer and both sides contain four hydrogen atoms and four oxygen atoms, so the equation is perfectly balanced. Because the product on the right contains molecular oxygen $$\mathrm{O_2},$$ the statement that the decomposition “gives di-oxygen” is absolutely correct. Hence, statement (a) is true.

Now we move to statement (b): “Like hydrogen peroxide, compounds such as $$\mathrm{KClO_3},\; \mathrm{Pb(NO_3)}$$ and $$\mathrm{NaNO_3}$$ when heated liberate di-oxygen.” We verify this one compound at a time.

For potassium chlorate we have the thermal decomposition formula (taught in school as the preparation of oxygen gas):

$$2\,\mathrm{KClO_3}\;\xrightarrow{\;\Delta\;}\;2\,\mathrm{KCl}\;+\;3\,\mathrm{O_2}.$$

Again, $$\mathrm{O_2}$$ appears on the product side, so $$\mathrm{KClO_3}$$ does indeed release di-oxygen.

For lead(II) nitrate—the correct formula is $$\mathrm{Pb(NO_3)_2}$$ but the idea is the same—the decomposition is written as

$$2\,\mathrm{Pb(NO_3)_2}\;\xrightarrow{\;\Delta\;}\;2\,\mathrm{PbO}\;+\;4\,\mathrm{NO_2}\;+\;\mathrm{O_2}.$$

Here also, gaseous $$\mathrm{O_2}$$ is liberated. Even if the original statement omitted one nitrate group and wrote $$\mathrm{Pb(NO_3)},$$ the intent was clearly the common laboratory salt that does evolve oxygen. Thus the spirit of the statement is correct.

For sodium nitrate we recall the heating equation

$$2\,\mathrm{NaNO_3}\;\xrightarrow{\;\Delta\;}\;2\,\mathrm{NaNO_2}\;+\;\mathrm{O_2}.$$

The coefficient “2” is chosen so that nitrogen and sodium atoms balance properly, and, crucially, $$\mathrm{O_2}$$ again appears. Therefore each of the cited salts behaves “like hydrogen peroxide” in the sense that they all give off molecular oxygen on heating. Consequently, statement (b) is true.

Next comes statement (c): “2-Ethylanthraquinone is useful for the industrial preparation of hydrogen peroxide.” The commercial anthraquinone process begins with 2-ethylanthraquinone (often abbreviated $$\mathrm{EAQ}$$). The molecule is first hydrogenated to the corresponding anthrahydroquinone and is then oxidized by air to regenerate the quinone and simultaneously produce $$\mathrm{H_2O_2}$$ in the working solution. Because this compound functions as a reversible carrier of hydrogen, it is indeed the cornerstone of modern large-scale $$\mathrm{H_2O_2}$$ manufacture. So statement (c) is unquestionably true.

Finally, consider statement (d): “Hydrogen peroxide is used for the manufacture of sodium perborate.” In industry sodium perborate is obtained by reacting hydrogen peroxide with sodium borate (or, more specifically, sodium metaborate) under suitable alkaline conditions. A simplified stoichiometric representation is

$$\mathrm{Na_2B_4O_7}\;+\;2\,\mathrm{NaOH}\;+\;4\,\mathrm{H_2O_2}\;\longrightarrow\;2\,\mathrm{NaBO_3\!\cdot\!4H_2O}\;+\;2\,\mathrm{H_2O}.$$

Because $$\mathrm{H_2O_2}$$ is consumed to introduce the peroxo linkage in the borate lattice, its use in the manufacture is genuine. Therefore statement (d) is also true.

We have now shown that statements (a), (b), (c) and (d) are all correct. Among the listed options, only Option A includes every one of these four statements.

Hence, the correct answer is Option A.