Mathematical proof of ΔQ=ΔH at constant external pressure
In an irreversible process, the total force per unit area exerted by the gas on the piston face is comprised of the local thermodynamic gas pressure at the piston face (as determined by the local gas density and temperature at the piston face) plus a viscous stress related to the rate of deformation of the gas in close proximity to the piston. So the thermodynamic pressure is not uniform throughout the gas and there are viscous stresses present.  By comparison, in a reversible process, the pressure is uniform throughout the gas and the viscous stresses are negligible.  In both cases, if the piston is massless, the force per unit area exerted by the gas on its side of the piston is exactly equal to $p_{ext}$, the force per unit area on the outside of the piston, and the work will be the same, as calculated from the force on either side of the piston.

If a frictionless piston has mass and the cylinder is horizontal, the piston will oscillate about its final position until the viscous stresses damp out its motion (i.e., its kinetic energy dissipates).  In the end, the final gas pressure will then be equal to $p_{ext}$, and, from a transient force and energy balance on the piston, one can show that the total work done by the gas will still integrate to $p_{ext}\Delta V$.

If the compression takes place irreveribly by, at time zero, suddenly increasing the external force per unit area from the initial gas pressure $p_0$ to the constant external force per unit area $p_{ext}$, the change in enthalpy of the gas from its initial state to its final state will not be equal to Q.  We will have $$W=p_{ext}(V_f-V_0)$$ and $$\Delta U = Q-W=Q-p_{ext}(V_f-V_o)$$so$$Q=\Delta U+p_{ext}(V_f-V_o)$$while$$\Delta H=\Delta U+(p_{ext}V_f-p_0V_0)$$