The variation of the parcel enthalpy with pressure during the interval of declining pressures over which the temperature remains constant at 0 C is still given by: $$dH=VdP\tag{1}$$Since the temperature of the bone dry air does not change, its contribution to dH is zero. Only the relative changes in the amounts of water vapor, liquid water, and ice determine dH. Let ##\omega_S##, ##\omega_L##, and ##\omega_V## represent the number of moles of ice, liquid water, and water vapor present in our parcel at any pressure P in the interval. Then, from a mass balance on the water, we have: $$\omega_S+\omega_L+\omega_V=\omega_{total}$$where ##\omega_{total}=\frac{p_s(279)}{84.5}## is the total number of moles of water in the parcel initially, a constant. From this it follows that the changes in the relative molar amounts are given by:
$$d\omega_S+d\omega_L+d\omega_V=0\tag{2}$$
The change in enthalpy of the overall parcel in the pressure interval over which the temperature remains constant at 0 C is then given by: $$dH=h_Sd\omega_S+h_Ld\omega_L+h_Vd\omega_V\tag{3}$$
where ##h_S##, ##h_L##, and ##h_V## are the molar enthalpies of saturated ice, saturated liquid water, and water vapor relative to an arbitrary reference state. If we eliminate ##d\omega_S## between Eqns. 2 and 3, we obtain: $$dH=(h_L-h_S)d\omega_L+(h_V-h_S)d\omega_V$$where ##(h_l-h_S)=L_F=6200\ J/mole## is the heat of fusion and ##(h_V-h_S)=L_S=(L+L_F)=45000 + 6200 = 51200\ J/mole## is the heat of sublimation. So we have: $$dH=L_Fd\omega_L+L_Sd\omega_V\tag{4}$$During the pressure interval where the temperature is constant, the mole fraction of water vapor in the parcel is given by $$\omega_V=\frac{p_s(273)}{P}$$So $$d\omega_V=-\frac{p_s(273)}{P^2}dP$$and thus our equation for the enthalpy change becomes:
$$L_Fd\omega_L=L_S\frac{p_s(273)}{P^2}dP+\frac{273R}{P}dP\tag{5}$$Because this equation is not a function of temperature, it can be integrated analytically to get the change in the moles of liquid water as a function of the decreasing pressure. It should be integrated from the pressure at which the temperature first hits 0C to the pressure at which the moles of liquid water has dropped to zero.
From your numerical solution, what is the pressure at which the temperature first hits 273 K? How many moles of liquid water are present in the parcel at this pressure?
