Denote the lengths of three sides of $\triangle ABC$ by $BC = a$, $CA = b$ and $AB = c$ respectively. No loss of generality, we can suppose $b \neq c$. We choose a rectangular coordinate system (see the figure), then we have $A(m, n)$, $B(0, 0)$, $C(a, 0)$ and $P(x, y)$.

Since $AF^2 - BF^2 = AP^2 - BP^2$, it follows that
$$AF^2 - (c - AF)^2 = AP^2 - BP^2.$$
Therefore,
$$2c \cdot AF - c^2 = (x-m)^2 + (y-n)^2 - x^2 - y^2,$$
and
$$AF = \frac{m^2 + n^2 - 2mx - 2ny}{2c} + \frac{c}{2}.$$

Similarly, we can compute $BD$ and $CE$ in terms of $x$ and $y$.

Since $AF + BD + CE = \frac{l}{2}$, we get
$$\frac{m^2+n^2-2mx-2ny}{2c} + \frac{c}{2} + x + \frac{2mx+2ny-m^2-n^2-2ax+a^2}{2b} + \frac{b}{2} = \frac{l}{2},$$
that is,
$$\left(\frac{m}{b} - \frac{a}{b} - \frac{m}{c} + 1\right)x + \left(\frac{n}{b} - \frac{n}{c}\right)y + \frac{a^2}{2b} + \frac{b+c}{2} + \frac{m^2+n^2}{2}\left(\frac{1}{c} - \frac{1}{b}\right) - \frac{l}{2} = 0.$$
Since $b \neq c$ and $n \neq 0$, point $P$ is on a fixed straight line. Since the condition $2(AF + BD + CE) = l$ is satisfied for both incenter and circumcenter, we complete the proof.