Adjoining figure shows the snapshot of two waves A and B at any time t. The equation for A is y = A $\sin (kx-\omega t-\varphi )$ and for B, it is y = A $\sin (kx-\omega t)$. It is clearly shown in the figure that wave A is ahead of B by a distance  $\varphi /k$ . The motion of a single point, i.e., y due to two waves individually, is not represented by

The equation for wave A can  be rewritten as 
  $$\begin{gathered} y=A\sin [kx-\omega t-\varphi ] \\ y=A\sin [k(x-\varphi /k)-\omega t] \\ or y=A\sin [kx-\omega (r+\frac{\varphi }{k})] \end{gathered}$$
While equation of wave B is  $y=A\sin (kx-\omega t)$ .
Comparing above equations, we can easily conclude that A is at a distance ahead of  $\varphi /k$  from B or wave A is ahead of B by a time difference of  $\varphi /\omega$ .so, b.  is the correct option 
Remember! In y versus t” ahead of means to the left of” while in y versus x ”ahead of means to the right of” if the wave travel in positive x-direction and vice-versa.