4. Given real numbers $x, y$ satisfy $\frac{x^{2}}{3}+y^{2}=1$. Then
$$
P=|2 x+y-4|+|4-x-2 y|
$$

the range of values for $P$ is . $\qquad$
4. $[2,14]$.

Let $x=\sqrt{3} \cos \theta, y=\sin \theta(\theta \in[0,2 \pi))$.
Then $2 x+y=2 \sqrt{3} \cos \theta+\sin \theta$
$$
\begin{array}{l}
=\sqrt{13} \sin (\theta+\alpha)<4, \\
x+2 y=\sqrt{3} \cos \theta+2 \sin \theta \\
=\sqrt{7} \sin (\theta+\beta)<4 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence } P=|2 x+y-4|+|4-x-2 y| \\
=(4-2 x-y)+(4-x-2 y) \\
=8-3(x+y) \\
=8-3(\sqrt{3} \cos \theta+\sin \theta) \\
=8-6 \sin \left(\theta+\frac{\pi}{3}\right) .
\end{array}
$$

Since $\theta \in[0,2 \pi)$, we get $2 \leqslant P \leqslant 14$.