Solution:
Let $AP = p$, $BP = q$, $CP = r$, $DP = s$; $AB = a$, $BC = b$, $CD = c$ and $DA = d$. Let $\angle APB = \angle CPD = \theta$. Then $\angle BPC = \angle DPA = \pi - \theta$. Let us also write $PE = h_1$, $PF = h_2$, $PG = h_3$ and $PH = h_4$.

Observe that
$$h_1 a = p q \sin \theta, \quad h_2 b = q r \sin \theta, \quad h_3 c = r s \sin \theta, \quad h_4 d = s p \sin \theta$$

Hence
$$\frac{1}{h_1} + \frac{1}{h_3} = \frac{1}{h_2} + \frac{1}{h_4}$$
is equivalent to
$$\frac{a}{p q} + \frac{c}{r s} = \frac{b}{q r} + \frac{d}{s p}$$

This is the same as
$$a r s + c p q = b s p + d q r$$

Thus we have to prove that $a + c = b + d$ if and only if $a r s + c p q = b s p + d q r$. Now we can write $a + c = b + d$ as
$$a^2 + c^2 + 2 a c = b^2 + d^2 + 2 b d$$

But we know that
$$\begin{aligned} & a^2 = p^2 + q^2 - 2 p q \cos \theta, \quad c^2 = r^2 + s^2 - 2 r s \cos \theta \\ & b^2 = q^2 + r^2 + 2 q r \cos \theta, \quad d^2 = p^2 + s^2 + 2 p s \cos \theta \end{aligned}$$

Hence $a + c = b + d$ is equivalent to
$$- p q \cos \theta - r s \cos \theta + a c = p s \cos \theta + q r \cos \theta + b d$$

Similarly, by squaring $a r s + c p q = b s p + d q r$ we can show that it is equivalent to
$$- p q \cos \theta - r s \cos \theta + a c = p s \cos \theta + q r \cos \theta + b d$$

We conclude that $a + c = b + d$ is equivalent to $c p q + a r s = b p s + d q r$. Hence $ABCD$ has an incircle if and only if
$$\frac{1}{h_1} + \frac{1}{h_3} = \frac{1}{h_2} + \frac{1}{h_4}$$