We have $\triangle ABC$ and we must find $PM$ 
Here we have $\triangle ABC$ and some information related:
$$\angle A=60^\circ$$
$$\angle B=45^\circ$$
$$AC=8$$
$$CM=MB$$
The vertical line $PM$ is perpendicular to $BC$.
So now we want to calculate $PB$.
$$PB=?$$
I have tried some different ways to calculate it. But it was unsuccesfully not true.
What I tried was to calculate $\angle C=75^\circ$ and dividing the   $\triangle ABC$ into 4 pieces and try to find the side $PM$.like this:

Do you have any ideas? Please explain your answer briefly.

With the information of the second picture, you can calculate $CH$ and $PS$.
If you know $PS$, you know $HM$. Than $HM + CH = MB$ 
And $MB = PM$
Knowing $PM$ and $MB$ you can calculate $PB$
