= Solution
Here $T$ is the absolute <temperature> and $k_B$ the <Boltzmann constant>. An <adsorption blob> contains $g\sim(\xi/b)^2$ <monomers>, so there are $N/g\sim N(b/\xi)^2$ such blobs. Confining a blob removes an amount of conformational <entropy> of order $k_B$; thus the positive confinement contribution to the <Helmholtz free energy> is
$$
F_{\mathrm{conf}}\simeq k_BT\frac Ng=k_BT N\left(\frac b\xi\right)^2.
$$
The statement about one blob's <entropy> is used as an <entropy> \i[loss] relative to the free chain, rather than as an absolute <entropy>.
In the assumed slab model, contacts occur within a monomer-scale distance $b$ of the substrate. The contact fraction is approximately $b/\xi$, giving $N_c\simeq Nb/\xi$. Multiplying this by the negative contact <energy> yields
$$
F_{\mathrm{contact}}=-\epsilon k_BT\frac{Nb}{\xi},\qquad
\boxed{F(\xi)=Nk_BT\left[\left(\frac b\xi\right)^2-\epsilon\frac b\xi\right].}
$$
The two contributions express the competition between loss of conformational <entropy> and gain of <adsorption> <energy>. The numerical coefficients are the ones stipulated by this blob approximation.
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