SIDE NOTE I'll start wy saying that your solubility product constants should be the other way around - the K_(sp) for lead (II) chloride should be 1.19 * 10^(-4) and the K_(sp) for silver chloride should be 1.79 * 10^(-10).
Since the other answer used them as they were given to you, I'll solve again in more detail using the correct solubility product constants.
Here's how you'd go about solving this problem. Take a look at the three equilibrium reactions you have to work with
color(blue)((1)): PbCl_2 rightleftharpoons Pb^(2+) + 2Cl^(-), K_3 = 1.19 * 10^(-4)
color(blue)((2)): AgCl rightleftharpoons Ag^(+) + Cl^(-), K_4 = 1.79 * 10^(-10)
color(blue)((3)): PbCl_2 + 2Ag^(+) rightleftharpoons 2AgCl + Pb^(2+)
Notice that equilibrium color(blue)((2)) has the ions on the products' side, but equilibrium color(blue)((3)) has 2 silver cations on the reactants' side. This means that you must reverse the second equilibrium and multiply it by 2.
2Ag^(+) + 2Cl^(-) rightleftharpoons 2AgCl
The solubility product constant will be
K_"1/4" = 1/K_4^2 = 1/(1.79)^2 * 10^20 = 3.12 * 10^19
Add the two equilibrium reactions to get
PbCl_2 + 2Ag^(+) + cancel(2Cl^(-)) rightleftharpoons Pb^(2+) + cancel(2Cl^(-)) + 2AgCl
When you add equilibrium reactions together, the resulting K_(sp) will be the product of the solubility product constants of the two separate reactions.
As a result, you'll get
K_"final" = K_3 * K_"1/4"
K_"final" = 1.19 * 10^(-4) * 3.12 * 10^(19) = color(green)(3.7 * 10^15)
