Understanding the derivation of Gibbs Free Energy
The surroundings are treated as a constant volume reservoir, so the first law of thermodynamics,
$$
\mathrm{d}U_\mathrm{surroundings} 
  = \mathrm{d}q_\mathrm{surroundings} 
  + \mathrm{d}w_\mathrm{surroundings},
$$
yields 
$$
\mathrm{d}U_\mathrm{surroundings} 
  = \mathrm{d}q_\mathrm{surroundings}
  \text{, or }
\Delta U_\mathrm{surroundings} 
  = q_\mathrm{surroundings}.
$$ 
Notice that $q_\mathrm{surroundings}$ is equal to a state function, so it behaves as one, that is, it is independent whether the process is reversible or not. If the process is reversible
$$
q_\mathrm{rev,surroundings} 
  = \Delta H_\mathrm{surroundings},
  \tag1\label{eq:rev}$$
if it is not reversible 
$$
q_\mathrm{surroundings} 
  = \Delta H_\mathrm{surroundings},
  \tag2\label{eq:irrev}$$
from \eqref{eq:rev} and \eqref{eq:irrev} follows
$$q_\mathrm{surroundings} = q_\mathrm{rev,surroundings}.$$
Therefore 
$$
\Delta S_\mathrm{surroundings} 
  = \frac{q_\mathrm{rev,surroundings}}{T},
$$ 
in particular 
$$q_\mathrm{rev,surroundings} = -q_\mathrm{sys},$$ 
then 
$$\Delta S_\mathrm{surroundings} = \frac{-q_\mathrm{sys}}{T}.$$
If the process occurs at constant pressure 
$$q_\mathrm{sys} = \Delta H_\mathrm{sys},$$ 
therefore 
$$\Delta S_\mathrm{surroundings}  = \frac{-\Delta H_\mathrm{sys}}{T}.$$