Adiabatic isobaric irreversible process
For an adiabatic isobaric process the work done by the system during a spontaneous expansion is $w= -p_{\text{ext}} \Delta V $.  Since the heat $q=0$ the work is also equal to the change in internal energy, ie $w= \Delta U$. The associated enthalpy change is then 

$$\begin{align} \Delta H &amp;= \Delta U + \Delta (pV) \\ &amp;= w + p_fV_f-p_iV_i \\ &amp;=  -p_{\text{ext}}\Delta V+ p_{\text{ext}} V_f-p_iV_i \\ &amp;=  p_{\text{ext}}V_i-p_iV_i\\&amp;=(p_{\text{ext}}-p_i)V_i \end{align}$$

In order for the expansion to proceed spontaneously the initial pressure of the system $p_i$ must exceed that of the surroundings $p_{\text{ext}}$, which means $\Delta H&lt;0$.

Since you are computing $H$, a property of the system, when computing the change in $H=U+pV$ you should use the pressure of the system, not of the surroundings. Before the expansion the pressure of the system (and surroundings) is $p_i$, while at the end of the expansion it is $p_f=p_\text{ext}$, since the system is then again in mechanical equilibrium with the surroundings. Note that previous to the expansion the system and surroundings are also in mechanical equilibrium. The pressure of the surroundings is then suddenly dropped to a lower constant value. The following figure illustrates the process: