To determine which element has the lowest first ionization enthalpy among the given options, we look at the atomic numbers:

Atomic number 32 corresponds to Germanium (Ge).

Atomic number 35 corresponds to Bromine (Br).

Atomic number 87 corresponds to Francium (Fr).

Atomic number 19 corresponds to Potassium (K).

The element with the lowest first ionization energy is Francium (Fr) with atomic number 87. This can be represented by its electron configuration: $\text{Fr} - [\text{Rn}] 7s^1$. Francium, being a part of the alkali metal group, has the most loosely bound outer electron compared to the other elements listed, resulting in a lower first ionization enthalpy.