$p(x)$ will be a multiple of $g(x)$ if $g(x)$ divides $p(x)$

Now, $\quad g(x)=2-3 x=0$ gives $x=\frac{2}{3}$

Remainder $=p\left(\frac{2}{3}\right)=\left(\frac{2}{3}\right)^{3}-\left(\frac{2}{3}\right)+1$

$=\frac{8}{27}-\frac{2}{3}+1=\frac{17}{27}$

since remainder $\neq 0,$ So, $p(x)$ is not a multiple of $g(x)$.