4. As shown in Figure 1, in the Cartesian coordinate system, the vertices $A, B$ of rectangle $A O B C$ have coordinates $A(0,4), B(4 \sqrt{3}, 0)$, respectively. Construct the symmetric point $P$ of point $A$ with respect to the line $y=k x (k>0)$. If $\triangle P O B$ is an isosceles triangle, then the coordinates of point $P$ are $\qquad$
4. $\left(\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{33}}{3}\right)$ or $(2 \sqrt{3}, 2)$.

From the problem, we know $O P=O A=4, O B=4 \sqrt{3}$.
Since $\triangle P O B$ is an isosceles triangle, we have
$B P=B O$ or $O P=P B$.
When $B P=B O=4 \sqrt{3}$,

As shown in Figure 3, draw $P F \perp O B$.
Let $P(x, y)$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
4^{2}=x^{2}+y^{2}, \\
(4 \sqrt{3})^{2}=(4 \sqrt{3}-x)^{2}+y^{2}
\end{array}\right. \\
\Rightarrow P\left(\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{33}}{3}\right) .
\end{array}
$$

Figure 3

When $O P=P B=4$, it is easy to see that,
$$
\begin{array}{l}
O F=F B=\frac{1}{2} O B=2 \sqrt{3} \\
\Rightarrow P F=\sqrt{O P^{2}-O F^{2}}=2 \\
\Rightarrow P(2 \sqrt{3}, 2) .
\end{array}
$$

Therefore, the required points are $P\left(\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{33}}{3}\right)$ or $P(2 \sqrt{3}, 2)$.