The oxime of tert-butyl phenyl ketone, $(\mathrm{CH}{}_3){}_3\mathrm{C}-\mathrm{C}(\mathrm{C}{}_6\mathrm{H}{}_5)=\mathrm{N}-\mathrm{OH}$, is drawn with the $\mathrm{OH}$ group on the same side of the $\mathrm{C}=\mathrm{N}$ bond as the phenyl ring. Its CIP descriptor and the reason are

On nitrogen, $\mathrm{OH}$ outranks the lone pair. On carbon, both tert-butyl and the ipso carbon of phenyl are $(\mathrm{C},\mathrm{C},\mathrm{C})$ (phenyl with a duplicate carbon). At the next sphere the phenyl branches are $\{\mathrm{C},\mathrm{C},\mathrm{H}\}$, $\{\mathrm{C},\mathrm{C},\mathrm{H}\}$ and a duplicate carbon, while tert-butyl branches are $\{\mathrm{H},\mathrm{H},\mathrm{H}\}$ three times. So phenyl outranks tert-butyl. $\mathrm{OH}$ is cis to phenyl, so the descriptor is $Z$.