1. Given wave

The incident wave is $y_1=a\sin(\omega t-kx).$

This is a wave traveling in the $+x$ direction.

2. Condition for a node at $x=0$

A node means the resultant displacement is always zero at that point. So at $x=0$, $y_1(0,t)+y_2(0,t)=0\quad \text{for all } t.$

Now, $y_1(0,t)=a\sin(\omega t).$

Hence the unknown wave must satisfy $y_2(0,t)=-a\sin(\omega t).$

3. Test each option at $x=0$

Option A

$y_2=a\sin(\omega t+kx)$ At $x=0$, $y_2(0,t)=a\sin(\omega t).$ So, $y_1+y_2=2a\sin(\omega t)\neq 0.$ Not correct.

Option B

$y_2=-a\sin(\omega t+kx)$ At $x=0$, $y_2(0,t)=-a\sin(\omega t).$ Thus, $y_1+y_2=0$ for all $t$ at $x=0$. So this gives a node at $x=0$.

Also, the resultant becomes $y=a\sin(\omega t-kx)-a\sin(\omega t+kx).$ Using $\sin C-\sin D=2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right),$ we get $y=-2a\sin(kx)\cos(\omega t).$ At $x=0$, $y=0,$ confirming a node.

Option C

$y_2=a\sin(\omega t-kx)$ At $x=0$, $y_2(0,t)=a\sin(\omega t).$ So resultant is not zero. Not correct.

Option D

$y_2=-a\sin(\omega t-kx)$ At $x=0$, $y_2(0,t)=-a\sin(\omega t).$ This does make the displacement zero at $x=0$. But this wave has the same propagation form as the first wave, so it is not the second oppositely traveling wave that forms a standing wave by meeting the first wave. The phrase "meets with another wave" implies a wave traveling in the opposite direction. Hence this is not the intended physical answer.

4. Conclusion

The unknown wave must be the oppositely traveling wave with a phase such that $x=0$ is a node: $\boxed{y=-a\sin(\omega t+kx)}.$

So the correct option is B.