jmc

To build $P(x),$ we start with the equation $x=\sqrt{2}+\sqrt{3}+\sqrt{5}$ and repeatedly rearrange and square the equation until all the terms have rational coefficients. First, we subtract $\sqrt{5}$ from both sides, giving $$x-\sqrt{5}=\sqrt{2}+\sqrt{3}.$$Then, squaring both sides, we have $$\begin{array}{rl}(x-\sqrt{5}{)}^2& =5+2\sqrt{6}\\ x^2-2x\sqrt{5}+5& =5+2\sqrt{6}\\ x^2-2x\sqrt{5}& =2\sqrt{6}.\end{array}$$Adding $2x\sqrt{5}$ to both sides and squaring again, we get $$\begin{array}{rl}{x}^2& =2x\sqrt{5}+2\sqrt{6}\\ x^4& =(2x\sqrt{5}+2\sqrt{6}{)}^2\\ x^4& =20x^2+8x\sqrt{30}+24.\end{array}$$To eliminate the last square root, we isolate it and square once more: $$\begin{array}{rl}{x}^4-20x^2-24& =8x\sqrt{30}\\ (x^4-20x^2-24{)}^2& =1920x^2.\end{array}$$Rewriting this equation as $$(x^4-20x^2-24{)}^2-1920x^2=0,$$we see that $P(x)=(x^4-20x^2-24{)}^2-1920x^2$ is the desired polynomial. Thus, $$\begin{array}{rl}P(1)& =(1-20-24{)}^2-1920\\ & =43^2-1920\\ & =\boxed{-71}.\end{array}$$