Actually, the equation does exist. It's 57.11. Ask yourself what Zload is on a shorted stub? Clearly, it is 0. This simplifies equation 57.11 to Zin = Zo * j * tan(Bl). Beta can be substituted via equation 57.5.
The really lame part is in the wording. The admittance at point X is 0.2 + 0.4j Siemens. Notice if that pesky 0.4j weren't there then the admittance would be 0.2 and, therefore, the impedance would be 50 ohms. So, the stub needs to be -0.4j Siemens. This is because admittances in parallel add. The -0.4j Siemens admittance from the stub is the same as saying 1/-0.4j = 25j ohm impedance. This is Zin on the stub. Now you have Zin = 25j going into the equation (simplified 57.11 above with Zo = 75 ohms). You can solve for l, which equals 40.9mm.
THAT SAID. There appears to be a typo that I missed when studying for the October exam. The solution contains a (-j) term in the sentence starting with "Analytically, the input...". It would be nice to get a second opinion on that, but their analysis proceeds as if it's positive.
Anyway, I can always write the problem up if need be and post it for clarification.