Eliminating $b$ and $c$ gives $(p^2-2)(a^2-pa+2)=0$. If $p\ne\pm\sqrt2$ then $a=\frac{p\pm\sqrt{p^2-8}}2$ and substituting into $c+\frac2a=p$ gives $c=a$. But we are given that $a,b,c$ are distinct. Hence we must have $p=\pm\sqrt2$. Then $2=p^2=\left(a+\frac2b\right)\left(b+\frac2c\right)=ab+\frac{2a}c+\frac4{bc}+2$ $\implies$ $0=abc+2\left(a+\frac2b\right)=abc+2p$.