This problem is tricky, but doable if you break down all your components. You will need to use properties of similar Triangles, properties of parallel and perpendicular lines, and some simple algebra. First, DRAW A DIAGRAM (seriously, you must do this).
To start, note that PMN is a right triangle. Draw a line perpendicular through point K that is perpendicular to PM and label the intersection as point Q. Observe that angle PQK is a right angle and that both triangles PKQ and PNM contain angle MPN. By AA, PKQ ~ (is similar to) PNM. This allows us to say something about the ratio of their side lengths.
Since the two triangles are similar, the following holds PK/PN = PQ/PM=KQ/MN. Since we were given that PK = (1/5)PN, we have the constant of proportionality that allows us to find the missing values.
KQ = (1/5)MN = (1/5)*20 = 4.
PQ = (1/5)PM = (1/5)*16 = 16/5.
In your diagram, make sure to show that PQ = 16/5 and QK = 4.
Now, draw a line through point K that is perpendicular to LB and label the intersection as point R. Then, draw a line through point P that is perpendicular to RK and label the intersection as point T. NOTE: We have now shown that LPKB is a composition of 3 figures: LPKB = LPTR + PTK + KRB.
Observe that since PQKT forms a rectangle so PT = QK = 4 and PQ = TK = 16/5.
Since LP = PT = 4, LPTR is a square and has an area of 16.
PTK is a right triangle so its area is ½ * PT * TK = ½ * 4 * 16/5 = 32/5.
BRK is a right triangle so its area is ½ * BR * RK = ½ * BR * (RT + TK). Since LPTR is a square, LR = RT = 4. Since LMNB is a square, LB = 20. Since LB = LR + RB, we know 20 = 4 + RB and thus RB = 16.
This means that the area of triangle BRK = ½ * BR * (RT + TK) = ½ * 16 * (4 + 16/5) = ½ * 16 * (36/5) = 288/5.
Thus, the area of LPKB = the summed areas of LPTR + PKT + BRK = 16 + 32/5 + 288/5 = 80.
So, the area is 80 cm^2.