Very short calculation using barycentric coordinates.

Let $X,Y \in DB$ be such that $AX$ and $BY$ are perpendicular to $DB$. Let $XA=YB=h, XD=a, YB=b, AB=c, CD=kc$. Moreover, let $\angle PAB=\angle DAC=x$ and $\angle PBE=\angle DBC=y$. We need to prove that $PA+PB \leq \dfrac{1}{\sqrt{1-k^2}} \cdot AB$, that is $\sin x + \sin y \leq \dfrac{1}{\sqrt{1-k^2}} \sin(x+y)$. Since $\sin(x+y)=\sin x \cos y+\sin y \cos x$, we are left to prove that $\sqrt{1-k^2}(\sin x + \sin y) \leq \sin x \cos y+\sin y \cos x$ We have the following small Claim: Claim: $\sin x=\dfrac{DC \cdot h}{AD \cdot AC}$ and $\sin y=\dfrac{DC \cdot h}{BD \cdot BC}$. Proof: We will only prove the first equality, as the proof for the second one is similar. Note that $\dfrac{AD \cdot AC \sin x}{2}=(ADC)=\dfrac{DC \cdot h}{2},$ hence we obtain the desired equality $\blacksquare$ By our Claim, we are equivalently left to prove that $\sqrt{1-k^2} \cdot (\dfrac{1}{AD \cdot AC}+\dfrac{1}{BD \cdot BC}) \leq \dfrac{\cos y}{AD \cdot AC}+\dfrac{\cos x}{BD \cdot BC},$ that is $\sqrt{1-k^2}(BD \cdot BC+AD \cdot AC) \leq BD \cdot BC\cos y+AD \cdot AC\cos x$ Now, we have our second Claim: Claim 2: $BD \cdot BC\cos y+AD \cdot AC\cos x=AD^2+BC^2+AB \cdot CD-CD^2$ Proof: Let $P \in AD$ and $Q \in BC$ be such that $CP$ is perpendicular to $AD$ and $DQ$ is perpendicular to $BC$. Note that $BD \cdot BC\cos y+AD \cdot AC \cos x=BQ \cdot BC+AD \cdot AP=$ $=BC^2+AD^2+QC \cdot BC+PD \cdot DA=BC^2+AD^2+DC \cdot CY+DC \cdot CX=$ $=AD^2+BC^2+(DY+CX) \cdot DC=AD^2+BC^2+AB \cdot CD-CD^2,$ as desired $\blacksquare$ Therefore, returning to the problem, we need to prove that $AD^2+BC^2+AB \cdot CD-CD^2 \geq (AD \cdot AC+BD \cdot BC)\sqrt{1-k^2},$ that is (switching to variables and heavy algebra) $2h^2+a^2+b^2+(k-k^2)c^2 \geq \sqrt{1-k^2}(\sqrt{(h^2+a^2)(h^2+(c-b)^2)}+\sqrt{(h^2+b^2)(h^2+(c-a)^2)})$ Note that we have the condition $a+b=c-kc$, as $a+b=XA+CY=XY-DC=AB-DC=c-kc$. By the Cauchy-Schwarz inequality, it suffices to prove that $(2h^2+a^2+b^2+(k-k^2)c^2)^2 \geq (1-k^2)((h^2+a^2)+(h^2+b^2))((h^2+(c-b)^2)+(h^2+(c-a)^2)),$ or equivalently (using $a+b=c-kc$) that $(2h^2+a^2+b^2+(k-k^2)c^2)^2 \geq (1-k^2)(2h^2+a^2+b^2)(2h^2+2kc^2+a^2+b^2)$ Put $2h^2+a^2+b^2=M$. Then, we are left to prove that $M(M+2kc^2)(1-k^2) \leq (M+(k-k^2)c^2)^2,$ which is equivalent to $(kM-(k-k^2)c^2)^2 \geq 0,$ which holds. Hence, we may conclude.