a: show that 2K = 3 ∫ P dV = 3 ∫ (P /ρ) dm

Not terribly difficult:
P = (1/3) ∫ n(p) pv d ³p ergs/cm³
KE per unit volume = ∫ n(p) pv/2 d ³p ergs/cm³ = 3P/2,
and dV = (1/&rho)dm .

b: write U and Ω in terms of M, R, and f for our two-layer star. Assume all the material is at the same temperature. Pretty easy, since everything can be derived from K. Now, find the Kelvin-Helmholtz time scale t_KH = R / (dR /dt) by assuming L = − dW /dt = constant. It should, of course, be t_KH = constant × GM ²/LR.

A bit harder. Our model assumes constant density inside fR, containing total mass M/2, and the rest just outside.
The potential Ω = − ∫_o^M (Gm/r) dm.
The inner half mass may be written
− ρ ²(16/3)π²G ∫_o^fRr ⁴dr = − (3/5)GM ²/(4fR).
The outer half mass provides an additional −GM ²/(4fR).
So, Ω = − (2/5)GM ²/fR.
Then, at least for ideal gasses, K = U = −Ω/2, and W = Ω/2.
L = − dW /dt = [(1/5)GM ²/fR ² ] dR /dt,
so t_KH = (1/5)GM ²/LfR