Entropy and specific heat capacity
It is not true that heat exchanged at constant pressure is always reversible.
But, if you want to determine the change in entropy from thermodynamic equilibrium state 1 at $(T_1,P)$ to state 2 at $(T_2,P)$, you need to forget entirely about the actual irreversible process path that took you from state 1 to state 2.  It is of no further use.  You instead need to focus exclusively on the two end states.  And, in order to apply the entropy equation, you need to devise a reversible path between these exact same two ends states.  Any reversible path will do, because they will all give the same value for the entropy change.
In the case of your example, you can choose a constant pressure reversible path for convenience. So, for such a path, from the first law we know that $\mathrm{d}H = \mathrm{d}q$. But, now how do we figure out a reversible path between the two states in which the system temperature is changing reversibly along the path?  If we contact the system with at constant temperature reservoir held at the final thermodynamic equilibrium temperature $T_2$, that path would not be reversible because there would be a finite temperature difference between the system and the reservoir over most of the path. But, now, suppose we contact the system, not with a single constant temperature reservoir, but with a sequence of constant temperature reservoirs, each at a slightly different temperature over the range $T_1$ to $T_2$. This would guarantee that, over the entire path, the system would differ only slightly from the reservoir it is currently in contact with. So, in this case, $\mathrm{d}q_\mathrm{rev} = \mathrm{d}H = C_p\,\mathrm{d}T$, where $\mathrm{d}T$ is the change in the system temperature over the present increment of heat exchanged. We use the equation
$$\mathrm{d}H = C_p\,\mathrm{d}T$$
because of the precise definition of constant pressure heat capacity in thermodynamics:
$$C_p\equiv\left(\frac{\partial H}{\partial T}\right)_P$$
So now, for the change in entropy, we have:
$$\Delta S = \int{\frac{\mathrm{d}q_\mathrm{rev}}{T}} = \int_{T_1}^{T_2}\frac{C_p \,\mathrm{d}T}{T}$$
So the key thing to remember here is that, if you want to determine the change in entropy for an irreversible process, you first need to determine the initial and final end states (say, using the first law of thermodynamics) and then devise a reversible path between these same two end states to calculate the integral of $\dfrac{\mathrm{d}q_\mathrm{rev}}{T}$ for that path.