The answer to this question is no. Let me explain why, the explanation uses two facts.

1) The complement to the collection of $4$ generic lines in $\mathbb CP^2$ is not $K(\pi,1)$.

Indeed, it is not hard to see that the fundamental group of this complement is $\mathbb Z^3$. At the same time the complement has homotopy type of a $CW$ complex of dimension $2$ and if it were $K(\pi, 1)$, it would zero cohomology in degree $3$, while $H^3(\mathbb Z^3)=\mathbb Z$.

2) There exist line arrangements in $\mathbb CP^2$ that contain $4$ generic lines, but at the same time their complement is $K(\pi,1)$. For example, you can take the arrangement given by $9$ lines $x=0$, $y=0$, $z=0$, $x=\pm y$, $y=\pm z$, $z=\pm x$. The generic four lines in it are $x=\pm y$, $x=\pm z$.

Proof. Taking the above arrangement we start to through away lines from it until we get the arrangement composed of $4$ generic lines. Hence at some point we will pass from an arrangement whose complement is $K(\pi,1)$ to the one whose complement is not $K(\pi,1)$. End.

In order to show that the complement to the above arrangement is $K(\pi,1)$ you can use, for example the theorem of Deligne, that tells that the complement to a complexification of a real simplicial arrangement is $K(\pi,1)$. (in our case real simplicial means that the $9$ lines are real and they cut $\mathbb RP^2$ in triangles.)

Here is the reference: Deligne, Pierre Les immeubles des groupes de tresses généralisés. (French) Invent. Math. 17 (1972)

The answer to this question is no. Let me explain why, the explanation uses two facts.

1) The complement to the collection of $4$ generic lines in $\mathbb CP^2$ is not $K(\pi,1)$.

Indeed, it is not hard to see that the fundamental group of this complement is $\mathbb Z^3$. At the same time the complement has homotopy type of a $CW$ complex of dimension $2$ and if it were $K(\pi, 1)$, it would zero cohomology in degree $3$, while $H^3(\mathbb Z^3)=\mathbb Z$.

2) There exist line arrangements in $\mathbb CP^2$ that contain $4$ generic lines, but at the same time their complement is $K(\pi,1)$. For example, you can take the arrangement given by $6$ lines $x=0$, $y=0$, $z=0$, $x=y$, $y=z$, $z=x$. The generic four lines in it are $x=0$, $y=0$, $x=z$, $y=z$.

Proof for No. Taking the above arrangement throw away from first the line $z=0$, second the line $x=y$. Hence at the first or second time we will pass from an arrangement whose complement is $K(\pi,1)$ to the one whose complement is not $K(\pi,1)$. End.

You can check that the complement to the above $6$ lines arrangement is $K(\pi,1)$ either directly, or you can use, for example the theorem of Deligne, that tells us that the complement to a complexification of a real simplicial arrangement is $K(\pi,1)$. (in our case real simplicial means that the $6$ lines are real and they cut $\mathbb RP^2$ in triangles.)

Here is the reference: Deligne, Pierre Les immeubles des groupes de tresses généralisés. (French) Invent. Math. 17 (1972)