For all antinodal positions we know that
[tex]\Delta \phi = 2N\pi[/tex]
now we also know the relation between phase difference and path difference
[tex]\Delta \phi = \frac{2\pi}{\lambda}\Delta x[/tex]
now we will have
[tex]2N\pi = \frac{2\pi}{\lambda}\Delta x[/tex]
now from above equation we will have
[tex]\Delta x = N\lambda[/tex]
now for the first anti node position
N = 1
[tex]\Delta x = \lambda[/tex]
Option D is correct