7. Let point $P$ move on the right branch of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ excluding the vertices, $E$ and $F$ be its left and right foci, respectively. Let $A$ be the excenter of $\triangle P E F$ in $\angle P E F$. Then the equation of the trajectory of point $A$ is $\qquad$
7. $\frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1(x>5)$.

From the problem, we know that the semi-major axis length $a=4$, and the semi-focal distance $c=5$.
Let $A(x, y), P\left(x_{0}, y_{0}\right)$. Then $x_{0}>4$.
Let $P A$ intersect the $x$-axis at point $B\left(x_{B}, y_{B}\right)$.
By the angle bisector property, we have
$\frac{|B A|}{|A P|}=\frac{|B E|}{|P E|}, \frac{|B A|}{|A P|}=\frac{|B F|}{|P F|}$.
Thus, $\frac{|B A|}{|A P|}=\frac{|B E|-|B F|}{|P E|-|P F|}=\frac{2 c}{2 a}=\frac{5}{4}$.
Therefore, point $A$ divides the segment $\overrightarrow{B P}$ in the ratio $\lambda=\frac{5}{4}$.
By the section formula, we get
$\left\{\begin{array}{l}x=\frac{x_{B}+\lambda x_{0}}{1+\lambda}=\frac{4 x_{B}+5 x_{0}}{9}, \\ y=\frac{y_{B}+\lambda y_{0}}{1+\lambda}=\frac{5 y_{0}}{9} .\end{array}\right.$
By the focal radius formula of the hyperbola, we get
$|P F|=e x_{0}-a=\frac{5}{4} x_{0}-4$.
And $|B F|=x_{B}-5$, thus,
$$
\frac{x_{B}-5}{\frac{5}{4} x_{0}-4}=\frac{5}{4} \Rightarrow x_{B}=\frac{25}{16} x_{0} .
$$

From the system of equations (1), we obtain
$$
x_{0}=\frac{4}{5} x, y_{0}=\frac{9}{5} y \text {. }
$$

Since $P\left(x_{0}, y_{0}\right)$ lies on the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, the equation of the locus of point $A$ is
$$
\frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1(x>5) .
$$