If $x+1$ is a factor of $p(x),$ then $p(-1)=0$

$(a)$ Let $p(x)=x^{3}+x^{2}-x+1$

$\therefore \quad p(-1)=(-1)^{3}+(-1)^{2}-(-1)+1$

$=-1+1+1+1=2 \neq 0$

So, $x+1$ is not a factor of $p(x)$

$(b)$ Let $p(x)=x^{3}+x^{2}+x+1$

$\therefore \quad p(-1)=(-1)^{3}+(-1)^{2}+(-1)+1$

$=-1+1-1+1=0$

$(c)$ Let $p(x)=x^{4}+x^{3}+x^{2}+1$

$\therefore \quad p(-1)=(-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)+1$

$=1-1+1+1=2 \neq 0$

$(d)$ Let $p(x)=x^{4}+3 x^{3}+3 x^{2}+x+1$

$\therefore \quad p(-1)=(-1)^{4}+3(-1)^{3}+3(-1)^{2}+(-1)+1$

$=1-3+3-1+1=1 \neq 0$

Hence, $x+1$ is a factor of $x^{3}+x^{2}+x+1$

So, $(b)$ is the correct answer.