Example 3 In $\triangle A B C$, $\angle C=90^{\circ}, B C=2$, $P$ is a point inside $\triangle A B C$ such that the minimum value of $P A+P B+P C$ is $2 \sqrt{7}$. Find the degree measure of $\angle A B C$.

Analysis: The key to this problem is to find the line segment when $P A+P B+P C$ reaches its minimum value. From the analysis of Example 2, if we construct an equilateral $\triangle A C D$ outside $\triangle A B C$ with $A C$ as one side, and connect $B D$, then the length of $B D$ is the minimum value of $P A+P B+P C$.
Solution: As shown in Figure 3, construct an equilateral $\triangle ACD$ outside $\triangle ABC$ with $AC$ as a side, and connect $BD$. Then the length of $BD$ is the minimum value of $PA + PB + PC$, i.e., $BD = 2\sqrt{7}$. Draw $DE \perp BC$ intersecting the extension of $BC$ at point $E$.
Let $AC = x$. Then $CD = x$.
In the right $\triangle CDE$,
$\angle DCE = 180^{\circ} - \angle ACD - \angle ACB = 30^{\circ}$.
Thus, $DE = \frac{1}{2} CD = \frac{1}{2} x$,
$EC = CD \cos \angle DCE = \frac{\sqrt{3}}{2} x$.
In the right $\triangle ADE$, $BE^2 + DE^2 = BD^2$, i.e., $\left(2 + \frac{\sqrt{3}}{2} x\right)^2 + \left(\frac{1}{2} x\right)^2 = (2\sqrt{7})^2$.
Solving this, we get $x = 2\sqrt{3}$.
In the right $\triangle ABC$,
$\tan \angle ABC = \frac{AC}{BC} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Thus, $\angle ABC = 60^{\circ}$.
Note: The three segments forming a "Y" shape can be transformed into a connected broken line by constructing an equilateral triangle.