Drop feet from $C$ on $BP$. Angle conditions give congruency and we are done.

Let $Q$ be the point on $AD$ such that $QC=QD$. Then the condition $\angle DBC+2\angle ADC=180^{\circ}$ implies $BCDQ$ is concyclic. Thus $\angle QCB=\angle ADB$ or $\angle QCP = \angle QDP$, which implies $PC=PD$. Now let $R$ be the point on ray $PA$ such that $AR=AP$. Hence $BCDR$ is cyclic, implying $PB=PR$ so we are done.

One can just angle chase and $PC=PD$, and through obtaining all angles in terms of one variable $x$ use sine rule on $\triangle APD $ and $\triangle BPC$.

$Let$ $\angle BAD=x^{\circ}$. $Simple$ $angle$ $chasing$ $gives$ $\angle ACB=2x$, $\angle CDP=\angle DCP=\frac{90-x}{2}$. $So,$ $PD=PC$ $In$ $\triangle APD$, $\sin x=\frac{AP}{PD}\implies AP=PD\sin x$ $In$ $\triangle PCB$, $by$ $Sine$ $rule,$ $\frac{\sin 2x}{\sin 90-x}=\frac{BP}{PC}\implies \frac{BP}{2}=PC\sin x$ $(As$ $\frac{\sin 2x}{\sin 90-x}=\frac{2\sin x\cos x}{\cos x}=2\sin x)$ $But,$ $PC=PD\implies PC\sin x=PD\sin x$ $Substituting$ $our$ $results,$ $\frac{BP}{2}=AP\implies 2AP=BP$

Why am I getting this: $\angle$ ADP = 90- $\angle$ APD = 90- $\angle$ BPC; as $\angle$ PCB = 2 $\angle$ ADP, $\implies$ $\angle$ PBC = 180-( $\angle$ BPC + $\angle$ PCB) =90- $\angle$ ADP $\implies$ 180-2( $\angle$ ADP + $\angle$ PDC) =90- $\angle$ ADP $\implies$ $\angle$ ADP +2 $\angle$ PDC =90 $\implies$ $\angle$ ACD = $\angle$ PDC $\implies$ $\angle$ DPC = 180-2 $\angle$ PDC $\implies$ $\angle$ APC = $\angle$ ADP + 180 -2 $\angle$ PDC =90. But A, P and C are on a straight line!!!

Sketch. Note that $PC=CB$ by angle chase. If $M$ is midpoint of $PB$, then triangles $APD$ and $PMC$ are congruent, because of $PD=PC$ (angle chase). So we're done.

Pretty simple by angle chasing , as far as I did. Denote $ \angle ADP=x$. So, $ \angle APD= \angle BPC = 90^{\circ}-x$ and $ \angle PCB= 2x$. This yields, $$ \angle PBC= 90^{\circ}-x \\ \implies \angle PBC= \angle BPC \\ \implies BC= PC $$Now, drawing a perpendicular from $C$ to point $M$ of $PB$ concludes $ \triangle PCM \sim \triangle PAD$. But we need to show their congruence.We can do this if we can prove $PD=PC$. How can we do that? Denote a point $Q$ on $AC$ such that $ \angle QDA=\angle ADC$ .Hence, $$\triangle QDA \cong \triangle ADC \implies QD=DC$$. Since $\triangle QDC \sim \triangle PDC$ , we can conclude $PD=PC$ . Thus we proved the congruence of $ \triangle PCM$ and $\triangle PAD$ which implies $PA=PM \implies PA=\frac{PB}{2} \implies 2PA=PB$ $\blacksquare$

Let CH be perpendicular to BP. ∠DBC + 2∠ADC = 180 ---> ∠DBC + 2∠ADB + 2∠PDC = ∠DBC + ∠BCP + 2∠PDC so ∠BPC = 2∠PDC and it means DPC is isosceles. ∠DAC = 90 = ∠DHC ---> AHCD is cyclic ---> ∠HCP = ∠PDA = ∠BCP/2 so CH is angle bisector of ∠BCP and it means BCP is isosceles. so we have BP = 2HP and now it's easy to show triangles HPC and APD are congruent so HP = AP. BP = 2HP = 2AP so we're Done.

$\color{magenta} \boxed{\textbf{SOLUTION}}$ From given condition, $$\angle DBC+2(\angle ADC)=180°\implies \angle DBC+2\angle ADB+2\angle BDC=180° \implies \angle DBC+\angle ACB+\angle BDC+\angle BDC=180° \implies 180°-\angle ACD+\angle BDC=180°$$$$\implies \angle ACD=\angle BDC$$ Let, $P'$ be the reflection of $P$ over $AD$ So, $$\angle BDP'=\angle PDP'=2\angle ADP=2\alpha$$From given condition, $$\angle BCP'=\angle ACB=2\angle ADB=2\alpha$$ So, $P'BCD$ is cyclic $$\implies \angle PBP'=\angle DBP'=\angle DCP'=\angle DCP=\angle CDP=\angle BDC=\angle BP'C=\angle PP'B \implies BP=PP'=2AP \blacksquare$$