Step 1: Identify the circuit function.
There is no resistor feeding the output back into either input, so this is not a negative-feedback amplifier stage. With an ideal (infinite gain) opamp and no feedback, output can only rail to $+15$ V or $-15$ V, whichever input is higher wins completely.
Step 2: Track the sign of $V(t)$ over one cycle.
The inverting input sits at 0 V (grounded), so the comparison reduces to the sign of $V(t)=2\sin(2000\pi t)$ at the non-inverting input. Checking a few instants: at $t=0$, $V=0$ and about to go positive; for small $t>0$, $\sin(2000\pi t)>0$ so $V_{out}=+15$ V; once $2000\pi t$ passes $\pi$, $\sin(\cdot)<0$ so $V_{out}$ flips to $-15$ V. This toggling repeats every time $\sin(2000\pi t)$ changes sign.
Step 3: Get the frequency, then the period.
Comparing $2000\pi t$ with the standard form $2\pi f t$ gives $f = \dfrac{2000\pi}{2\pi} = 1000$ Hz. So the period is
\[ T = \frac{1}{f} = \frac{1}{1000\ \text{Hz}} = 1\ \text{ms} \]
Since the output only ever sits at one of the two rails, its peak-to-peak swing is just the rail separation:
\[ V_{pp} = 15\ \text{V} - (-15\ \text{V}) = 30\ \text{V} \]
Step 4: Eliminate the wrong shapes and amplitudes.
A sine-shaped output (options B, C) would need the opamp working in its linear region with feedback, which is not present here, so both are ruled out on shape alone. A 4 V swing (option D) is far too small for a rail-to-rail comparator running off $\pm15$ V supplies.
Final Answer:
The comparator toggles between $+15$ V and $-15$ V at the input's zero crossings, giving a square wave of 30 V peak-to-peak and 1 ms period.
\[ \boxed{V_{pp} = 30\ \text{V},\ T = 1\ \text{ms}} \]