(N/A) $P$ $(Z=15)$: Electronic configuration is $[Ne] \, 3s^2 \, 3p^3$. The $3p$ subshell has $3$ electrons in separate orbitals,so there are $3$ unpaired electrons.
$(b)$ $Si$ $(Z=14)$: Electronic configuration is $[Ne] \, 3s^2 \, 3p^2$. The $3p$ subshell has $2$ electrons in separate orbitals,so there are $2$ unpaired electrons.
$(c)$ $Cr$ $(Z=24)$: Electronic configuration is $[Ar] \, 3d^5 \, 4s^1$. All $5$ electrons in $3d$ and $1$ electron in $4s$ are unpaired,totaling $6$ unpaired electrons.
$(d)$ $Fe$ $(Z=26)$: Electronic configuration is $[Ar] \, 3d^6 \, 4s^2$. In $3d^6$,one orbital is paired and four are unpaired,totaling $4$ unpaired electrons.
$(e)$ $Kr$ $(Z=36)$: Electronic configuration is $[Ar] \, 3d^{10} \, 4s^2 \, 4p^6$. All orbitals are completely filled,so there are $0$ unpaired electrons.