NiCl_ 2 \ P(C_ 2 H_ 5 )_ 2 (C_ 6 H_ 5 )\ _ 2 exhibits temperature...

In both states (paramagnetic and diamagnetic) of the given complex, $\mathrm{Ni}$ exists as $\mathrm{Ni}^{2+}$ whose electronic configuration is $[\mathrm{Ar}] 3 d^{8} 4 s^{0}$.
In the above paramagnetic state the geometry of the complex is $s p^{3}$ giving tetrahedral geometry. The diamagnetic state is achieved by pairing of electrons in $3 d$ orbital.
Thus, the geometry of the complex will be $d s p^{2}$ giving square planar geometry.