Let $X\equiv AC\cap BD$, $\Gamma_1$ and $\Gamma_3$ be the circumcircles of triangles $BCD$ and $DAB$ respectively. Clearly $A'$ and $C'$ lie on line $AC$. The power of $X$ with respect to $\Gamma_1$ is $A'X\cdot XC$ and the power of $X$ with respect to $\Gamma_3$ is $AX\cdot XC'$. But $X$ lies on the radical axis of those two circumferences so these quantities are equal; in particular we get: $A'X/C'X=AX/CX$. Similarly $B'X/D'X=BX/DX$. Since $ABCD$ is a trapezoid then $A'X/C'X=B'X/D'X$, which means that triangles $A'XB'$ and $C'XD'$ are homotetic with a negative factor which implies that $A'B'$ is parallel to $C'D'$.

As $P$ lies on the radical axis of $(ABC)$ and $(ADC)$, $$PB \cdot PD_1 = PD \cdot PB_1 \iff \frac{PD_1}{PB_1} = \frac{PD}{PB}.$$Similarly, $\frac{PC_1}{PA_1} = \frac{PC}{PA}$, so $$\frac{PD_1}{PB_1} = \frac{PD}{PB} = \frac{PA}{PC} = \frac{PA_1}{PC_1},$$and $\overline{A_1D_1} \parallel \overline{B_1C_1}$, as needed.