Show that $f_{k}\longrightarrow f$ both weakly in $L^{p}$ and a.e. implies $|f_{k}|^{p}-|f|^{p}-|f_{k}-f|^{p}$ converges to $0$ in distribution

(a) According to your own computation \begin{align*} \int_{-\pi}^{\pi}(|f_{k}|^{2}-|f|^{2}-|f_{k}-f|^{2})\phi dx &=\int_{-\pi}^{\pi}(2f_{k}f-2f^{2})\phi dx\\ &=2\int_{-\pi}^{\pi}(f_{k}-f)\, (f\phi)\, dx. \end{align*} Since $f\phi\in L^2(-\pi,\pi)$, the weak convergence implies that the integral on the right-hand side tends to 0 as $k\to\infty$.

(b) As a counterexample you could take $f_k(x):=\frac{1}{\sqrt{|x|}}(1-\chi_{[-\frac{1}{k},\frac{1}{k}]}(x))$ and $p=\frac{3}{2}$. Observe that $f_k$ converges weakly (strongly even) towards $f(x):=\frac{1}{\sqrt{|x|}}$in $L^{\frac{3}{2}}(-\pi,\pi)$. However, $|f|^2$ is not locally integrable and therefore not a distribution.