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

Now, $g(x)=2 x+1$ give $x=-\frac{1}{2}$

Remainder $=p\left(-\frac{1}{2}\right)=2\left(\frac{-1}{2}\right)^{3}-11\left(\frac{-1}{2}\right)^{2}-4\left(\frac{-1}{2}\right)+5$

$=2\left(\frac{-1}{8}\right)-11\left(\frac{1}{4}\right)+2+5=\frac{-1}{4}-\frac{11}{4}+7$

$=\frac{-1-11+28}{4}=\frac{16}{4}=4$

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