4. Given $a b c \neq 0$, and $\frac{a+b}{c}=\frac{b+c}{a}=$ $\frac{c+a}{b}=p$. Then, the line $y=p x+p$ must pass through
(A) the first and second quadrants
(B) the second and third quadrants
(C) the third and fourth quadrants
(D) the first and fourth quadrants
4. (B).

From the conditions, we get $\left\{\begin{array}{l}a+b=p c, \\ b+c=p a, \\ a+c=p b .\end{array}\right.$
Adding the three equations, we get $2(a+b+c)=p(a+b+c)$.
$\therefore p=2$ or $a+b+c=0$.
When $p=2$, $y=2 x+2$. Thus, the line passes through the first, second, and third quadrants.

When $a+b+c=0$, let's assume $a+b=-c$, then, $p=\frac{a+b}{c}=-1 .(c \neq 0)$
$\therefore y=-x-1$, thus, the line passes through the second, third, and fourth quadrants.
In summary, the line must pass through the second and third quadrants. Therefore, the answer is (B).