jmc

We perform polynomial division with $P(x)$ as the dividend and $Q(x)$ as the divisor, giving $$\begin{array}{rl}P(x)=x^6-x^5-x^3-x^2-x& =(x^2+1)(x^4-x^3-x^2+1)+(x^2-x+1)\\ & =(x^2+1)Q(x)+(x^2-x+1).\end{array}$$Thus, if $z$ is a root of $Q(x)=0,$ then the expression for $P(z)$ is especially simple, because $$\begin{array}{rl}P(z)& =\cancel{(z^2+1)Q(z)}+(z^2-z+1)\\ & =z^2-z+1.\end{array}$$It follows that $$\sum _{i=1}^{4}P(z_i)=\sum _{i=1}^4(z_i^2-z_i+1).$$By Vieta's formulas, ${\sum }_{i=1}^4{z}_i=1,$ and $$\sum _{i=1}^4{z}_i^2={\left(\sum _{i=1}^4{z}_i\right)}^2-2\sum _{1\le i<j\le 4}{z}_i{z}_j=1^2-2(-1)=3.$$Therefore, $$\sum _{i=1}^{4}P(z_i)=3-1+4=\boxed{6}.$$