= Solution
For the outer $ns,np$ <valence electrons>, \b[<Slater's rules> predict a modest initial rise down group 15, followed by approximately constant values]. Write the <effective nuclear charge> as $Z_{\rm eff}=Z-S$, where $S$ is the <shielding constant>. Each atom has $ns^2np^3$, so the other four electrons in the outer group contribute $4(0.35)=1.40$. Electrons in shell $n-1$ contribute $0.85$ each, and those deeper contribute $1.00$ each.<Atomic Shielding Constants by John C. Slater>[1930, shielding rules on printed pp. 58–59]{ref}
Applying those weights gives the following original calculations:
$$
\begin{array}{c|r|r|r}
&Z&S&Z_{\rm eff}(ns,np)\\\hline
\mathrm N&7&1.40+2(0.85)=3.10&3.90\\
\mathrm P&15&1.40+8(0.85)+2=10.20&4.80\\
\mathrm{As}&33&1.40+18(0.85)+10=26.70&6.30\\
\mathrm{Sb}&51&1.40+18(0.85)+28=44.70&6.30\\
\mathrm{Bi}&83&1.40+18(0.85)+60=76.70&6.30
\end{array}
$$
The increasing nuclear <electric charge> is largely offset by additional inner-electron shielding; once the preceding shell contains $18$ electrons, these simple weights give a plateau. The $n-1$ shell includes its d electrons. This trend concerns outer-shell values, not every orbital: a $1s$ electron is shielded only by its partner, so $Z_{\rm eff}(1s)=Z-0.30$ rises strongly down the group. The larger principal shell also means nearly constant outer $Z_{\rm eff}$ does not imply nearly constant atomic size or binding energy.
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