How do I go about selecting the resistors for pnp common emitter amplifier?
Assumption:- With an operating frequency of 10kHz your 10uF capacitors are effectively short circuits at AC compared to other impedances in the circuit. 

First thing to do, and most difficult is to decide what your collector current is going to be. It has to be high enough to drive your output impedance in parallel with the dc setting resistor Rc, with Rl at about 1.7k with a voltage swing of 2Vpp you need a current swing of about 1.2 ma in the load. I am going to pick a DC current of 5ma in the collector this gives me the room to drive the load and is well within the transistors best operating region.

Next what is Rc? We usually try to set Collector voltage at half the supply to give maximum positive and negative excursion. This would give Rc=6.5V/5ma = 1.3k.

Now lets look at the emitter resistors. For good DC stability we want the emitter at around Supply-1V. The emitter current is about the same as the collector current so lets set Re+Re1=1V/5ma=200ohms. 

We need the AC gain to be 13. As you pointed out the AC gain is approximately (Rl in parallel with RC)/Re. As @LvW has kindly pointed out the approximation is due to the transconductance of the transistor. This has the effect of reducing the effective gain. There are two ways of handing this do some complex calculations and come up with a number that is only valid for a certain set of conditions or throw in some extra gain  simulate or build and adjust as needed. Lets take the second approach assume  (Rl in parallel with RC)/Re = 15. This puts Re as 75ohms so Re1 is 125ohms.

The average base current is going to be 5ma/hfe(min) which is going to be around 63uA for this transistor.

As a rule of thumb the current through the base voltage divider should be about ten times this so that the base current does not mess with the voltage division to much. So lets say this current is 650uA.

The base voltage needs to be about  Ve-.7 =supply-1.7 so R2 is 1.7/650ua=2.6k AND R1 is (13-1.7)/650uA=10.38k.

This approach has involved lots of "abouts" on the whole in practical design it is a much simpler and faster approach than working out every last theoretical equation. Build it or simulate it then correct it you will probably have to do that anyway.