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.

Guide:
I would construct line $PC$ and show that $\angle APC$ is $90^\circ$. 
After that I can use trigonometry to obtain $PC$ and I can use trigonometry one more time to obtain the length of $PM$.
Edit:
$PC$ is the line connecting $P$ to $C$. Notice that we have $PM = CM$. Then, we can compute $\angle ACP = 75^\circ - 45^\circ=30^\circ$. Hence $\angle APC = 90^\circ$. Hence $PC= AC \cos 30^\circ$.
