Likewise since enthalpy $H$ is a state function it is thus path-independent and so in the absence of non-PV work, it has to follow that $\Delta H= \Delta q$ applies equally to all isobaric paths whether they are reversible or not.

@ManRow Hello ManRow. I don't have much time to engage because I have to do some exams. You are not correct. I assure you that $Q = \Delta H$ is only true for isobaric and mechanically reversible processes. I attach to you the pages of the following book that explain this, in here. It is underlined in red. If you want, you can read its Chapter 2, and come back for next discussions.

We can ask @ChetMiller. He is a chemical engineer as me, and has read Smith, Van Ness, and Abbott one million times. What is your answer Chet? If the process is isobaric but irreversible, can we say that $Q = \Delta H$?

@ManRow Here you have another proof, in Chapter 2 of Atkins' book in Physical Chemistry. Pay attention to the underlined lines. If the process is irreversible, can you say $\mathrm{d}w = -p\,\mathrm{d}V$? No. If you can't say that, you can't replace that expression and reach $Q = \Delta H$. I await your answer to this discussion. Otherwise, I will flag these comments for words such as 'doesn't make sense' and not backing up your claims.

If you have an isobaric process where $V$ is decreased, you must also decrease $T$ in order to still ensure that $PV = nRT$ (since at constant pressure $P\Delta V = nR \Delta T$, so $V_2 < V_1 \Rightarrow T_2 < T_1$).

Also when internal and external pressures are the same then it always follows that $dW = -pdV$ regardless of whether the process is reversible or not. Here is an answer on the Physics-SE detailing this: physics.stackexchange.com/a/760921/168695

@ManRow I agree that the image is wrong, in the sense that the volume seems to "decrease". Unfortunately, I decreased the volume to put that text above that says $Q_\mathrm{surr{ = -Q$. Fortunately, in science and technology, we have authorative sources which I have cited to you. This helps us in the sense that if I write nonsense, they won't publish it. That's why I cited you two books. In that post, he/she statesit, doesn't details it. I won't go for further discussion. Thanks.

I think your confusion lies in the nature of irreversible paths. An isobaric path (reversible or not) requires only that the external pressure $P_{ext}$ is kept constant. A reversible isobaric path further requires that $P_{int} = P_{ext}$ at all times; so, if at some moment an isobaric path does not meet this requirement, then it must therefore be an irreversible path instead.

Likewise, for any reversible isobaric process, you can contrive an "irreversible" one with same start and end points (so that $\Delta H_{irr}=\Delta H_{rev}$) and adjust it to perform to the same total amount of work as well. Consequently, since $\Delta H_{irr} = \Delta H_{rev}$ and $\Delta W_{irr} = \Delta W_{rev}$, it must follow that $\Delta Q_{irr} = \Delta Q_{rev}$ and so $\Delta H_{irr} = \Delta H_{rev} = \Delta Q_{rev} = \Delta Q_{irr} \Rightarrow \Delta H_{irr} = \Delta Q_{irr}$.

This is simple to construct -- for example, if we must slowly add a total amount of $Q$ heat to expand some piston reversibly isobarically, then let's instead just add $Q/10$ to that piston instantaneously (so $P_{int}$ quickly (but temporarily!) exceeds $P_{ext}$), then apply the remaining $9Q/10$ slowly, so that we "eventually" makes it up to same desired height (same as if the process were 100% reversible at all times).

In short, given a reversible isobaric path, you can artificially contrive an "irreversible" isobaric one for the same start and end points with the same $\Delta W$, thereby making $\Delta H_{irr} = \Delta H_{rev} = \Delta Q_{rev} = \Delta Q_{irr}$.

Notice the use of "big delta $\Delta$" here -- an irreversible process needs only to have $dH\ne\delta q$ at some moments or so, but can still have the "overall" $\Delta H = \Delta Q$ (big delta) hold when looking at the process as a "whole"...

Or, on the other hand, perhaps I'm misinterpreting irreversibility? It seems the nature of irreversibility mandates the "creation of new entropy that wasn't there before". So, for example, in the case of isobaric expansion, it must always be that you have to supply more $Q_{irr}$ than $Q_{rev}$ in order to perform the same $W_{rev}$ for the same start and end points/conditions.

Thus, it would be impossible for an irreversible process to have the same $W$ and $Q$ as a reversible one for the same start and end points due to said entropy generation.