B

根据题意得，$\alpha +\beta =p$①，$\alpha \beta =q$②；
$\alpha ^{2}+\beta ^{2}=p$③，$\alpha ^{2}\beta ^{2}=q$④.
由②④可得$\alpha ^{2}\beta ^{2}-\alpha \beta =0$，
解之得$\alpha \beta =1$或$0$
由①③可得$\alpha ^{2}+\beta ^{2}=\left(\alpha +\beta \right)^{2}-2\alpha \beta =p^{2}-2q=p$，
即$p^{2}-p-2q=0$，
当$q=0$时，$p^{2}-p=0$，
解之得，$p=0$或$p=1$，
即$\left\{\begin{array}{l}{p}_{1}=0\\{q}_{1}=0\end{array}\right.$，$\left\{\begin{array}{l}{p}_{2}=1\\{q}_{2}=0\end{array}\right.$，
把它们代入原方程的$\triangle$中可知符合题意.
当$q=1$时，$p^{2}-p-2=0$，
解之得，$p=-1$或$2$，
即$\left\{\begin{array}{l}{p}_{3}=2\\{q}_{3}=1\end{array}\right.$，$\left\{\begin{array}{l}{p}_{4}=-1\\{q}_{4}=1\end{array}\right.$，
把它们代入原方程的$\triangle$中可知$\left\{\begin{array}{l}{p}_{4}=-1\\{q}_{4}=1\end{array}\right.$不合题意舍去，
所以数对$\left(p,q\right)$的个数是$3$对.
故选：$B$.