$∆H$ when non-expansion work is done I have learnt that at constant pressure, enthalpy change $∆H=q$ provided no non-expansion work is done. Why is this the case? What will be relationship of $∆H$ with $q$ if non-expansion work is done? Again, Gibbs free energy measures the maximum non expansion work that can be done by a system. Will $∆G$ appear in relationship of $∆H$ and $q$ when non expansion work is done? Every source I have come across only mentions the case when no non expansion work is done.
 A: For chemical reactions, the change in enthalpy $H$ (enthalpy of reaction) is, as you already said, defined as the heat $Q$ of reaction (assuming pressure $p$ is constant). This leads to
$$\Delta_rH=\Delta U+p\cdot V\tag{1}$$
(the index $_r$ is for "reaction")
This equation might seem a bit random at first, but it does make sense: We know that the change in internal Energy $U$ is
$$\Delta U=Q+W\tag{2}$$
where $W$ is the work done on or by the system. If expansion work $W_V=-p\cdot\Delta V$ is done by the system, the change in internal energy will be
$$\Delta U=Q-p\cdot\Delta V\tag{3}$$
Putting this equation $(3)$ into equation $(1)$, we get
$$\Delta_rH=Q\tag{4}$$
which is exactly how enthalpy is defined. However, we have assumed until now that only expansion work $W_V$ is done1. If we have some other work $W_a$ done on the system, we get
$$W_{\text{total}}=W_V+W_a\tag{5}$$
Plugging this into equation $(2)$ and then into equation $(1)$, one gets
$$\boxed{\Delta_rH=Q+W_a}\tag{6}$$
where $W_a$ is the sum of all non-expansion work done on the system. If this work is done by the system, we need to consider it negative.

1 I use the index $_V$ for "Volume" indicating that the system's volume changes during expansion work.
A: 
I have learnt that at constant pressure, enthalpy change $∆H=q$
provided no non-expansion work is done.

What you refer to as "expansion work" is normally referred to as boundary (expansion or compression) work, or $pdV$ work, which is applicable to closed systems (no mass transfer involved). The other possible type of work (non expansion work) is flow work, or $Vdp$ work, which is applicable to open systems. This is the work involved with pushing fluid into or out of the boundaries of a control volume.  So for non-expansion work, $dp\ne 0$.

What will be relationship of $∆H$ with $q$ if non-expansion work is
done?

That will depend on the type of process involved.
For example, in the case of an adiabatic ($Q=0$) steam turbine, $\Delta H=W$ where $W$ is the shaft work of the turbine and $\Delta H$ is the difference between the entering and exiting enthalpies for the turbine control volume. Those enthalpies combine the entering and exiting flow work and internal energies of the steam.
Hope this helps.
