4. Given that $P$ is a point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, $F_{1}$ is its left focus, and $Q$ is on $P F_{1}$ such that
$$
\overrightarrow{O Q}=\frac{1}{2}\left(\overrightarrow{O P}+\overrightarrow{O F_{1}}\right),|\overrightarrow{O Q}|=3 \text {. }
$$

Then the distance from point $P$ to the left directrix of the ellipse is
4.5.

From the problem, we know that $Q$ is the midpoint of side $F_{1} P$. Let the right focus of the ellipse be $F_{2}$, and connect $P F_{2}$. Then
$$
\begin{array}{l}
|O Q|=\frac{1}{2}\left|P F_{2}\right|=3 \\
\Rightarrow\left|P F_{2}\right|=6 \Rightarrow\left|P F_{1}\right|=4 .
\end{array}
$$

Let the distance from point $P$ to the left directrix be $d$. Therefore,
$$
\frac{\left|P F_{1}\right|}{d}=e=\frac{c}{a}=\frac{\sqrt{a^{2}-b^{2}}}{a}=\frac{4}{5} \text {. }
$$

Thus, $d=5$.