Compound that is both paramagnetic and coloured is

 ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$  oxidation state $\mathrm{Cr}$= +6

${\mathrm{Cr}}^{6+}$ is [Ar]3d⁰

The oxidation state of $V$ is +4 in ${\mathrm{VOSO}}_4$

${\mathrm{V}}^{4+}$ is $3{\mathrm{d}}^1$

${\mathrm{K}}_3\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4\right]$ 

Oxidation state of $\mathrm{Cu}$ = +1

${\mathrm{Cu}}^{1+}$ is Ar[3d]¹⁰

$\left[{\mathrm{TiCl}}_6\right].{\left({\mathrm{NH}}_4\right)}_2$

${\mathrm{Ti}}^{4+}$ configuration is $3{\mathrm{d}}^2$

Has one unpaired d electron so it is paramagnetic and coloured.