What is the measurement of the $\measuredangle PBN$ in the figure below? For reference: In figure M, N and L are points of tangency. calculate the $\measuredangle PBN$. ABD is triangle rectangle.

My progress;
Draw BE $\implies \measuredangle BEN = 45^\circ=\measuredangle EMB\\ $
E = incenter implies AE, BE e DE are angle bissectors
I drew some auxiliary lines but I still haven't seen the solution

 A: Observe that by assumption $AI$ and $BD$ are both perpendicular to $AB$ and are thus parallel to each other. Hence, $\angle \, AIL = \angle \, DNL$. But, by the tangency property, $DN = DL$ so the triangle $DLN$ is isosceles and $\angle\, DNL = \angle \, DLN = \angle\, ALI$ so $$\angle\, AIL = \angle\, ALI$$ which means that triangle $ALI$ is isosceles and if you add to this fact the tangency property, you get $$AI = AL = AM$$ which means that $AMI$ is a right-angled isosceles triangle, so $$\angle\, AMP = \angle\, AMI = 45^{\circ}$$ But $EM \, \perp \, AB$ so $$\angle \, EMP = 45^{\circ}$$ In the incircle, $EM = EP$ so $EMP$ is isosceles with $$\angle \, EPM = \angle\, EMP = 45^{\circ}$$ which means that $$\angle \, MEP = 90^{\circ}$$ however, $EMBN$ is a square so $$\angle \, PEN = \angle \, PEM + \angle \, MEN = 90^{\circ} + 90^{\circ} = 180^{\circ}$$ which means that $P, E, N$ are colinear and in fact the incenter $E$ is the midpoint of the diameter $PN$. Since $BD$ is tangent to the incircle at the point $N$ and $PN$ is a diameter, $$\angle\, BNP = 90^{\circ}$$ Consequently, the triangle $BNP$ is right-angled with $$\frac{PN}{BN} = \frac{2 EN}{EN} = 2$$
However, $$\tan\Big(\angle \, PBN\Big) = \frac{PN}{BN} = 2 $$ which means that $$\angle \, PBN = \arctan\big(2\big) = 63.435$$
A: 
$\triangle LDN$ is isosceles. So $\angle ALI = \angle DLN = 90^\circ - \frac{\angle D}{2}$
$\angle LAI = \angle D, ~$ so $~\angle AIL = 180^\circ - (\angle LAI + \angle ALI)$
$= 90^\circ - \frac{\angle D}{2} = \angle ALI$
That leads $AI = AL = AM$
Now $\triangle IAM$ is isosceles with $\angle AMI = \angle AIM = 45^\circ$
Also $\angle PME = \angle EPM = 45^\circ$
So, $\angle PEM = 90^\circ$.
As $\angle MEN = 90^0$, $P, E, N$ are collinear with $PN = 2 BN$.
($BNEM$ is a square)
So finally you have a right triangle $\triangle PBN$ with perpendicular sides in ratio $1:2$ which is a well known right  triangle.
So, $\angle PBN \approx 63.5^\circ$
