7. Given the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$, respectively, and $O$ is the center of the ellipse $\Gamma$. $P$ is a point on the ellipse $\Gamma$. The excircle of $\triangle P F_{1} F_{2}$ inside $\angle P F_{1} F_{2}$, denoted as $\odot I$, is tangent to the $x$-axis at point $A$. A line through point $F_{2}$ perpendicular to line $P I$ intersects at point $B$. If $e$ is the eccentricity of the ellipse $\Gamma$, then $|O B|=$ $\qquad$
7. $a$.

As shown in Figure 3, let the points of tangency of $\odot I$ with $P F_{1}$ and $P F_{2}$ be $M$ and $N$ respectively; extend $F_{2} B$ to intersect line $P F_{1}$ at point $C$.
Then $|P M|=|P N|,\left|F_{2} N\right|=\left|F_{2} A\right|$
$$
\begin{aligned}
\Rightarrow & 2\left|F_{1} A\right|=\left|F_{1} M\right|+\left|F_{1} A\right| \\
& =\left|F_{1} P\right|+|P M|+\left|F_{1} F_{2}\right|+\left|F_{2} A\right| \\
& =\left|F_{1} P\right|+\left|F_{1} F_{2}\right|+|P N|+\left|F_{2} N\right| \\
& =\left|F_{1} P\right|+\left|F_{1} F_{2}\right|+\left|P F_{2}\right| \\
& =2 a+2 c \\
\Rightarrow & \left|F_{1} A\right|=a+c
\end{aligned}
$$
$\Rightarrow A$ is the right vertex of the ellipse.
Since $P I$ bisects $\angle C P F_{2}$ and $F_{2} B \perp P I$
$\Rightarrow \triangle P F_{2} C$ is an isosceles triangle
$\Rightarrow B$ is the midpoint of $F_{2} C$.
Also, $O$ is the midpoint of $F_{1} F_{2}$, thus, $O B \perp \frac{1}{2} F_{1} C$. Therefore, $|O B|=\frac{1}{2}\left|F_{1} C\right|=\frac{1}{2}\left(\left|F_{1} P\right|+|P C|\right)$ $=\frac{1}{2}\left(\left|F_{1} P\right|+\left|P F_{2}\right|\right)=\frac{1}{2} \times 2 a=a$.