= Solution
Give the dual $M^*=\operatorname{Hom}_k(M,k)$ the contragredient action $(g\lambda)(m)=\lambda(g^{-1}m)$. If $\delta_h$ is the functional dual to the basis element $h\in G$, then
$$
g\delta_h=\delta_{gh}.
$$
Thus the map $h\mapsto\delta_h$ extends to a $kG$-isomorphism
$$
\boxed{kG\cong(kG)^*.}
$$
This is also the left-module form of the fact that a <group algebra is a symmetric algebra>.
If $P$ is finitely generated and projective, it is a direct summand of $(kG)^r$. Dualizing makes $P^*$ a direct summand of $((kG)^*)^r\cong(kG)^r$, so $P^*$ is projective. The converse follows by dualizing again and using $P\cong P^{**}$.
For any finite-dimensional algebra $A$, the module $A^*$ is injective because
$$
\operatorname{Hom}_A(-,A^*)\cong\operatorname{Hom}_k(-,k)
$$
is exact. Since $kG\cong(kG)^*$, free $kG$-modules are injective, and so are their projective direct summands. Conversely, duality sends injectives to projectives, so every finite-dimensional injective is projective. Hence <projective modules over a finite group algebra are injective>.
Finally, let $P$ be indecomposable projective. It is also an indecomposable injective. Its nonzero <socle> contains a simple module $S$, and the <injective hull> $E(S)$ is a direct summand of $P$. Indecomposability forces $P=E(S)$. Since $S$ is essential in its injective hull, every simple submodule of $P$ equals $S$. Therefore
$$
\boxed{\operatorname{Soc}(P)\text{ is simple}.}
$$
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