For the circuit to behave as a phase splitter, you need Rc = Re, because then the gain as CE amplifier will be -Rc/Re = -1.
The design starts with this requirement: Ic = 1mA. Also, Vcc = 20V. This for itself doesn't determine the amplifier: you can design an infinite number of phase splitters with Ic = 1mA at this voltage. You still have a design choice remaining: where does the collector voltage set when there is not signal (ie, the quiescent voltage). You can put this voltage almost anywhere from 10V to 20V, and that position, together with Ic=1mA, will determine Re = Rc.
The books does a very smart choice: since a phase splitter drives two equal (in amplitude) waves, one at the emitter and other at the collector, the 20V rail is divided in two zones: from 0 to 10V, where the emitter wave will swing, and from 10V to 20V, where the collector wave will swing. So when quiescent, the emitter voltage should be at 5V (middle of its zone), and the collector should be at 15V (middle of its zone).
Now it's time to use Ohm's Law: if Ic = 1mA, and the quiescent collector is at 15V, that means that Rc drops 5V, so: Rc * Ic = 5 -> Rc * 1e-3 = 5 -> Rc = 5k, nearest typical value Rc = 4.7K. The very same applies to Re which, anyway, must be equal to Rc.
Now that we have Re, and we know the quiescent emitter is at 5V, we deduce that the quiescent base is at 5+0.6 = 5.6V (assuming the transistor is active, which is a safe assumption, though it is always best to check for it after finishing the design). From Vb = 5.6, and Ib = Ic /beta, one uses that the base divider must be 'stiff enough' to drive the transistor, meaning that Rb1 || Rb2 = Re*beta/10. This gives us an equation. Another equation comes from the fact that Vcc * Rb1 / (Rb1 + Rb2) = Vbe (the base divider gives the base voltage computed before). We solve that easy system of equations and we get Rb1 and Rb2. Finally check if the solution you got is consistent with an active transistor (sometimes there are surprises, especially at very high beta or temperature). And you are done.