Hydrogen atom is excited from ground state to another state with principal quantum number equal to $$4$$. Then the number of spectral lines in the emission spectra will be

When a hydrogen atom is excited to the level with principal quantum number $$n=4$$, the electron can later jump from this level (or any intermediate level reached during the cascade) to any lower-energy level. Each jump produces one spectral line.

For a given highest level $$n$$, the total number of distinct pairs of energy levels is obtained by choosing any two different levels out of the $$n$$ available levels $$\{1,2,3,\ldots ,n\}$$. The number of such pairs is the combination

$$N = {}^{n}C_{2} = \frac{n(n-1)}{2}$$

Here $$n=4$$, so

$$N = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$$

Listing them explicitly confirms the count:
$$4 \rightarrow 3,\; 4 \rightarrow 2,\; 4 \rightarrow 1,\; 3 \rightarrow 2,\; 3 \rightarrow 1,\; 2 \rightarrow 1$$ — a total of $$6$$ spectral lines.

Option D which is: $$6$$