At first we shall prove following lemma:
Lemma. If point $A'$ is antipode of vertex $A$ in circumcircle of triangle $ABC$, then
$$S_{ABC} - S_{A'BC} = \frac{1}{2} |BC|^2 \cot \angle A,$$
where $S_{XYZ}$ denotes the area of triangle $XYZ$.
Proof. Let $H$ be the orthocenter of triangle $ABC$. Notice that
$$S_{ABC} - S_{A'BC} = S_{ABC} - S_{HBC} = \frac{1}{2} BC \cdot AH = \frac{1}{2} |BC|^2 \cot \angle A.$$

Now let $AB = a$, $AD = b$, $\angle APB = \angle P_1$, $\angle BPC = \angle P_2$, $\angle CPD = \angle P_3$ and $\angle DPA = \angle P_4$. From the lemma, it's easy to see that
$$2S_{ABCD} - S_{XYZT} = \frac{a^2}{2} (\cot \angle P_1 + \cot \angle P_3) + \frac{b^2}{2} (\cot \angle P_2 + \cot \angle P_4) \\ = \frac{a^2}{2} \left( \frac{\sin (\angle P_1 + \angle P_3)}{\sin \angle P_1 \cdot \sin \angle P_3} \right) + \frac{b^2}{2} \left( \frac{\sin (\angle P_2 + \angle P_4)}{\sin \angle P_2 \cdot \sin \angle P_4} \right)$$
Without loss of generality, we can assume $\angle P_1 + \angle P_3 > 180^\circ$ (If $\angle P_1 + \angle P_3 = 180^\circ$ we are done) so $\sin(\angle P_1 + \angle P_3) < 0$. Let points $K, L, M$ and $N$ be the foot of the perpendicular lines from $P$ to $AB, BC, CD$ and $DA$, respectively. We need to prove that
$$\Leftrightarrow \frac{a^2}{2} \left( \frac{\sin (\angle P_1 + \angle P_3)}{\sin \angle P_1 \cdot \sin \angle P_3} \right) + \frac{b^2}{2} \left( \frac{\sin (\angle P_2 + \angle P_4)}{\sin \angle P_2 \cdot \sin \angle P_4} \right) < 0$$
$$\Leftrightarrow \frac{a^2}{\sin \angle P_1 \cdot \sin \angle P_3} > \frac{b^2}{\sin \angle P_2 \cdot \sin \angle P_4}$$
$$\Leftrightarrow \frac{AP}{\sin \angle ABP} \cdot \frac{CP}{\sin \angle CDP} > \frac{AP}{\sin \angle ADP} \cdot \frac{CP}{\sin \angle CBP}$$

$\Leftrightarrow \sin \angle$ CBP $\cdot \sin \angle$ ADP > $\sin \angle$ ABP $\cdot \sin \angle$ CDP

$$\Leftrightarrow BP \cdot \sin \angle CBP \cdot DP \cdot \sin \angle ADP > BP \cdot \sin \angle ABP \cdot DP \cdot \sin \angle CDP$$
$$\Leftrightarrow PL \cdot PN > PK \cdot PM$$
$$\Leftrightarrow \angle LKN + \angle LMN > \angle KLM + \angle KNM$$
$$\Leftrightarrow \angle APB + \angle CPD > \angle APD + \angle BPC$$
The latest assertion is true, so the proof is completed. ■