According to Cayley Hamilton theorem: Every Square matrix satisfies its own characteristic equation.

So$, P=\lambda I$

Given that $\lambda_{1}=0,\lambda_{2}=0.5,\lambda_{3}=3$

$P^{2}+2P+I=(\lambda I)^{2}+2(\lambda I)+I=\lambda ^{2}+2\lambda +1$

Put $\lambda=0$

$\lambda^{2}+2\lambda +1 = 0+0+1=1$

Put $\lambda=0.5$

$\lambda^{2}+2\lambda_{3} +1=(0.5)^{2}+2(0.5)+1=0.25+1+1=2.25$

Put $\lambda=3$

$\lambda^{2}+2\lambda +1=(3)^{2}+2(3)+1=9+6+1=16$