1.) Draw a picture (aka- free body diagram) to help visualize the problem.

2.) Break up the forces into necessary components (in green). **Note:** I have chosen the coordinate system, as indicated in the picture, to simplify the problem.

3.) Newton's Second Law (for x):

$\displaystyle \sum F_{x}=ma_{x} $

$\displaystyle \sum F_{x}=0 $

$\displaystyle F_{p,x}-F_{g,x}-f_{k}=0 $

$\displaystyle F_{p,x}=F_{g,x}+f_{k} $

$\displaystyle F_{p}\cos(18^{\circ})=F_{g}\sin(30^{\circ})+\mu_{k }n $

4.) Newton's Second Law (for y):

$\displaystyle \sum F_{y}=ma_{y} $

$\displaystyle \sum F_{y}=0 $

$\displaystyle n+F_{p,y}-F_{g,y}=0 $

$\displaystyle n=F_{g,y}-F_{p,y} $

$\displaystyle n=F_{g}\cos(30^{\circ})-F_{p}\sin(30^{\circ}) $

5. Substitute back into the equation derived for $\displaystyle \sum F_{x} $:

$\displaystyle F_{p}\cos(18^{\circ})=F_{g}\sin(30^{\circ})+\mu_{k }[F_{g}\cos(30^{\circ})-F_{p}\sin(30^{\circ})] $

6. Solve for $\displaystyle F_{p} $:

$\displaystyle F_{p}=F_{g}[\frac{\sin(30^{\circ})+\mu_{k}\cos(30^{\circ})}{ \cos(18^{\circ})+\mu_{k}\sin(18^{\circ})}]=mg[\frac{\sin(30^{\circ})+\mu_{k}\cos(30^{\circ})}{ \cos(18^{\circ})+\mu_{k}\sin(18^{\circ})}] $

Hopefully that will clear things up! Good luck!