The formation of  ${\mathrm{PH}}_4^{+}$ is difficult compared to ${\mathrm{NH}}_4^{+}$ because

There is one lone pair of electrons in NH₃ and its directional because it presents in the hybrid orbital. Also, N has no vacant d-orbitals, therefore the electron cloud is concentrated over N atom and easily available for proton. Therefore, NH₃ easily forms of ${\mathrm{NH}}_4^{+}$.

${\mathrm{NH}}_3+{\mathrm{H}}^{+}\to {\mathrm{NH}}_4^{+}$

In PH₃, lone pair is present in purely-s orbital and its non-directional. As a result, its overlapping tendency is appreciably reduced than the lone pair present in the hybrid orbital. Therefore, it is difficult for PH₃ to form ${\mathrm{PH}}_4^{+}.$