The heat content of a system is equal to the enthalpy only for a system that is at constant what?

Only for a system at constant pressure... By definition, #q = nDeltabarH = DeltaH# at constant pressure, where #DeltabarH# is the molar enthalpy in #"J/mol"#.

We can begin from the first law of thermodynamics

#DeltaU = q + w#,

where #q# and #w# are the heat flow and work, respectively,

and the relation of the internal energy #U# to the enthalpy #H#,

#DeltaH = DeltaU + Delta(PV)#

where #P# and #V# are pressure and volume.

When one plugs in the expression for #DeltaU#:

#DeltaH = q + w + Delta(PV)#

The common convention is that the work is defined by #w = -PDeltaV# when work is with respect to the system (i.e. negatively-signed when the system does work on the surroundings by expanding, with #DeltaV > 0#).

Furthermore, we apply the product rule from calculus (plus a bit extra) to see that #Delta(PV) = PDeltaV + VDeltaP + DeltaPDeltaV#. Therefore:

#DeltaH = q - cancel(PDeltaV + PDeltaV) + VDeltaP + DeltaPDeltaV#

As a result, the change in enthalpy is related to the heat flow as:

#color(blue)(DeltaH = q + VDeltaP + DeltaPDeltaV)#

Clearly, when the pressure is constant, we can see that #VDeltaP = 0#. We designate this as

#DeltaH = q_P#,

if both quantities are in #"J"#. Multiplying the left by #n/n#, where #n# is the mols, gives the first equation presented,

#color(blue)(nDeltabarH = q_P)#