The four quantum numbers for the outermost electron of potassium can be determined by first understanding the electronic configuration of potassium (atomic number 19). The electron configuration is as follows:

$$ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 $$

The outermost electron resides in the 4s orbital.

The four quantum numbers are as follows:

Principal quantum number ($n$): This indicates the energy level or shell number of the electron and for the 4s orbital, $n = 4$.

Azimuthal quantum number ($l$): This indicates the subshell or shape of the orbital. For an $s$ orbital, $l = 0$.

Magnetic quantum number ($m$): This indicates the orientation of the orbital in space. Since there is only one orientation for $s$ orbitals, $m = 0$.

Spin quantum number ($s$): This can be either $+\frac{1}{2}$ or $-\frac{1}{2}$. Typically, the first electron in an orbital has a spin of $+\frac{1}{2}$.

Thus, the four quantum numbers for the outermost electron of potassium are:

$$n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2}$$

This matches with Option B:

$$\mathrm{n}=4, l=0, \mathrm{~m}=0, s=+\frac{1}{2}$$