Applicability of Enthalpy-Heat relation for a non-isobaric process
Disclaimer: The below discussion only holds to be completely true for reversible processes and for ideal gases.
I think the problem is you're thinking of it the wrong way, that is, you first think that $ q= nC_p \Delta T$ and for constant pressure process $q_p= \Delta H$ , and hence, $ \Delta H = nC_p \Delta T$
In actuality, it is the other way around. Similar to how internal energy is always $nC_v \Delta T$, the enthalpy is always $nC_p \Delta T$ for an ideal gas. A simple proof is shown below:
$$ \Delta H = \Delta U + \Delta (PV)$$
For $n$ fixed moles of an ideal gas,
$$ U = n C_v \Delta T$$
and,
$$ \Delta (PV) = nR \Delta T$$
Hence,
$$ \Delta H =n\Delta T [  R + C_v]$$
Now, from Mayers relation,
$$ C_p  = C_v + R$$
And hence,
$$ \Delta H = n \Delta T C_p$$
So, as seen above, if you can accept that $\Delta U= nC_v \Delta T$ for an ideal gas always then you must be able to accept a similar statement for enthalpy

Regarding state variables:
Yes, there are indeed many states for an ideal gas which correspond to the same enthalpy since enthalpy is a function of the only temperature. The state variables of pressure and volume are irrelevant for enthalpy calculations unless you don't have the final and initial temperatures. In such a case, you can use the ideal gas law to find expression for the temperature at different states.

Note:
A discussion of proving $q_p =\Delta H$ can be found in this post
You may think the above proof is circular reasoning since regularly Mayers relation is proven by using $ \Delta H = nC_p \Delta T$. However, there exists another way to prove it by considering path functions. Have a look at the first twenty minutes of this lecture by MIT OCW.