1. Initial condition: ice and water coexist
   At $P=1\,\text{atm}$ and $T=273.15\,\text{K}$, ice and water are in equilibrium.

2. Effect of pressure on melting point of ice
   For the solid-liquid equilibrium, $\frac{dT}{dP}=\frac{T\,\Delta V}{\Delta H}$ where $\Delta V = V_{\text{liquid}}-V_{\text{solid}}$.

   For water:

   • ice is less dense than liquid water,
   • so $V_{\text{solid}} > V_{\text{liquid}}$,
   • hence $\Delta V = V_{\text{liquid}}-V_{\text{solid}}<0$

   Therefore, $\frac{dT}{dP}<0$

   This means: increasing pressure lowers the melting point of ice.

3. Now pressure is doubled, temperature kept constant at $273.15\,\text{K}$
   Since the melting point decreases when pressure increases, at the new higher pressure the melting point becomes less than $273.15\,\text{K}$.

   So at $273.15\,\text{K}$, the system is now at a temperature above the new melting point, hence ice tends to melt.

4. Which phase is favored?
   Higher pressure favors the phase with smaller volume. Liquid water has smaller volume than ice, so pressure converts ice into water.

   Thus, the amount of ice decreases, and if pressure increase is sufficient, the solid phase disappears completely.

5. Check options

   • A: Liquid phase disappears completely. False. Pressure favors liquid, not solid.
   • B: The amount of ice decreases. True.
   • C: The solid phase (ice) disappears completely. True for the new equilibrium at $2\,\text{atm}$ and $273.15\,\text{K}$, since the melting point is depressed below $273.15\,\text{K}$.
   • D: Volume of system increases. False. Converting ice to water decreases volume.

6. About single-correct format
   Physically, both B and C are consistent, but in standard exam intent, since ice-water coexistence at $273.15\,\text{K}$ is destroyed and liquid is the stable phase at higher pressure, the expected answer is C.

Final answer: $\boxed{\text{C}}$