First consider $P \neq 0$. Let $x_1,x_2, \dots ,x_6$ be the desired numbers (if they exist). Note that since the numbers are pairwise distinct, so two consecutive triplets cannot follow the same condition (otherwise $x_1+x_2+x_3=x_2+x_3+x_4$ or $x_1x_2x_3=x_2x_3x_4$ gives $x_1=x_4$, a contradiction). Then we have (WLOG) $$x_1+x_2+x_3=x_3+x_4+x_5=x_5+x_6+x_1=S \text{ and } x_2x_3x_4=x_4x_5x_6=x_6x_1x_2=P$$This gives $x_1+x_2=x_4+x_5$ and $x_1x_2=x_4x_5$. Squaring the first, and using the second relation, one easily gets $x_1-x_2=\pm(x_4-x_5)$. But then we have either $x_1=x_4$ or $x_1=x_5$, a contradiction. Now, suppose $P=0$. Note that we cannot have $x_{i-1}+x_i+x_{i+1}=x_i+x_{i+1}+x_{i+2}=S$; else $x_{i-1}=x_{i+2}$ (indices taken modulo $6$). This means that atleast one of the $x_i$'s is zero. WLOG take $x_1=0$. Then atleast one out of $x_2x_3x_4,x_3x_4x_5$ and $x_4x_5x_6$ must equal $P=0$, which means that another $x_i$ is zero, a contradiction. Thus, no such numbers exist.