Step 1: What a node really means.
A node is a point that never moves. So at $x=0$ the total displacement $y=y_1+y_2$ must stay zero at every instant of time. The first wave is $y_1=a\cos(kx-\omega t)$.
Step 2: Write the first wave at the node.
Put $x=0$ in the given wave. We get $y_1=a\cos(-\omega t)=a\cos(\omega t)$, because cosine is an even function.
Step 3: Demand zero total at the node.
For a permanent node we need $y_1+y_2=0$ at $x=0$ for all $t$. So $y_2$ at $x=0$ must equal $-a\cos(\omega t)$.
Step 4: Test the option that points to a reflected wave.
A wave moving in the opposite direction has the form $\cos(kx+\omega t)$. Try $y_2=-a\cos(kx+\omega t)$. At $x=0$ this gives $-a\cos(\omega t)$, which is exactly what we needed.
Step 5: Confirm it forms a standing wave.
Adding the two, \[ y=a\cos(kx-\omega t)-a\cos(kx+\omega t) \] and using the difference formula, \[ y=2a\sin(kx)\sin(\omega t). \]
Step 6: Final check.
At $x=0$, $\sin(0)=0$, so $y=0$ always. A genuine node sits at the origin, just as the question wants.
So the correct second wave is \[ \boxed{y=-a\cos(kx+\omega t)} \]