4. Given the quadratic function $f(x)=a x^{2}+b x+c$ whose graph is shown in Figure 1. Let
$$
\begin{aligned}
p= & |a-b+c|+ \\
& |2 a+b|, \\
q= & |a+b+c|+ \\
& |2 a-b| .
\end{aligned}
$$

Then ( ).
(A) $p>q$
(B) $p=q$
(C) $p<q$
(D) The relationship between $p$ and $q$ cannot be determined
4.C.

From Fig. 1, we know that $f(0)=c=0, f(1)=a+b+c>0$, then
$$
a+b>0, f(-1)=a-b+c=a-b1$, we get
$$
2 a+b>0 \text {. }
$$

And $2 a-b=a+(a-b)<0$, so,
$$
\begin{array}{l}
p=-(a-b)+(2 a+b)=a+2 b, \\
q=(a+b)-(2 a-b)=-a+2 b .
\end{array}
$$

Since $p-q=(a+2 b)-(-a+2 b)=2 a<0$, hence $p<q$.