The compound(s) which react(s) with NH$$_3$$ to give boron nitride (BN) is(are)

When a substance containing boron is heated with excess ammonia, formation of boron nitride BN is possible only when the overall reaction can supply the elements in the 1 : 1 atomic ratio B : N and simultaneously remove all other atoms (H, O, etc.) as stable volatile molecules such as $$H_2$$ or $$H_2O$$.

Option A B (elemental boron)
Elemental boron does not react with $$NH_3$$ even at red heat; it forms BN only with $$N_2$$ at temperatures above $$2000\;^{\circ}\!{\rm C}$$. Hence BN is not obtained from the reaction of boron with ammonia. Option A is therefore incorrect.

Option B $$B_2H_6$$ (diborane)
Diborane first forms a Lewis-adduct with ammonia:
$$B_2H_6 + 2NH_3 \;\longrightarrow\; \bigl[B_2H_6\cdot 2NH_3\bigr]$$
On heating the adduct strongly (≈1100 K) hydrogen is expelled and a very stable lattice of boron nitride is produced:
$$\bigl[B_2H_6\cdot 2NH_3\bigr] \;\xrightarrow[\;1100\text{ K}\;]{}\; 2BN + 6H_2$$
Thus $$B_2H_6$$ definitely yields BN with ammonia. Option B is correct.

Option C $$B_2O_3$$ (boric oxide)
Boric oxide reacts with dry ammonia at elevated temperatures (≈800-1000 K) according to
$$B_2O_3 + 2NH_3 \;\longrightarrow\; 2BN + 3H_2O$$
The water formed escapes as vapour, leaving a white solid of boron nitride. Hence $$B_2O_3$$ is also a precursor of BN with ammonia. Option C is correct.

Option D $$HBF_4$$ (fluoroboric acid)
Fluoroboric acid is already fully coordinated by four fluoride ions around boron. With ammonia it simply undergoes neutralisation to give ammonium fluoro-borate: $$HBF_4 + NH_3 \rightarrow NH_4BF_4$$. No B-N bond formation or elimination of side-products that could convert it to BN is observed. Therefore it does not give BN. Option D is incorrect.

Hence the compounds that react with $$NH_3$$ to give boron nitride are:
Option B (diborane, $$B_2H_6$$) and Option C (boric oxide, $$B_2O_3$$).

Final Answer: Option B, Option C