B

Given two wavelengths: \( \lambda_1 = 6300 \) Å and \( \lambda_2 = 5600 \) Å. The bright fringe of the first wavelength corresponds to the fourth bright fringe, so \( m_1 = 4 \).

For the dark fringe of the second wavelength, we have the formula for dark fringes in YDSE which is given by:
\[ y_d = \frac{(n + 0.5) \lambda D}{d} \]
where \( n \) is the order of the dark fringe, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

The bright fringe for \( \lambda_1 \) is given by:
\( y_b = \frac{m_1 \lambda_1 D}{d} \)
Setting the two equations equal as the conditions for the bright fringe of \( 6300 \) Å and dark fringe of \( 5600 \) Å coincide, we can use the respective wavelengths:

\[ \frac{4 \cdot 6300 D}{d} = \frac{(n + 0.5) \cdot 5600 D}{d} \]
When simplifying, we can cancel \( D \) and \( d \):
\[ 4 \cdot 6300 = (n + 0.5) \cdot 5600 \]
This leads to:
\[ n + 0.5 = \frac{4 \cdot 6300}{5600} \]
Calculating this yields:
\[ n + 0.5 = 4.5 \rightarrow n = 4 \]
Thus, the value of \( n \) equals \( 4 \), corresponding to option B.