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For , $k\in \mathbb N$ , let $$P_k(t)=C_k\left(\frac{1+\cos t}{2}\right)^k \text{ , } t\in [-\pi,\pi]$$where $C_k\in \mathbb R$ is choosen in such a way that $$\frac{1}{2\pi}\int_{-\pi}^{\pi}P_k(t)\,dt=1.$$

Then prove that $\displaystyle C_k \le \frac{\pi}{2}(k+1)$ for all $k\in \Bbb N$.

We have , $P_k(t)=C_k\cos^{2k}\left(t/2\right)$. Then , from $$\frac{1}{2\pi}\int_{-\pi}^{\pi}P_k(t)\,dt=1.$$ $$\implies 2C_k\int_0^{\pi/2}\cos^{2k}(z)\,dz=\pi$$ $$\implies C_k=\frac{\pi.k!}{\sqrt{\pi}.\Gamma \left(\frac{2k+1}{2}\right)}$$ Then , how I can proceed to complete the proof ?

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You may conclude through Gautschi's inequality or the log-convexity of the $\Gamma$ function that follows from the Bohr-Mollereup theorem. You are requesting that $$ C_k \int_{-\pi}^{\pi}\left(\frac{1+\cos x}{2}\right)^k\,dx = 2\pi $$ or $$ C_k = \left[\frac{1}{4^k}\binom{2k}{k}\right]^{-1} $$ so it is enough to show that $$ \frac{2}{\pi}\leq \frac{k+1}{4^k}\binom{2k}{k} $$ where the RHS is related with the reciprocal Wallis product.

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$$ C_k=\dfrac{\sqrt{\pi} k!}{\Gamma (k+\frac12)}=\dfrac{\sqrt{\pi}\: k(k-1)(k-2)\dots 1}{\sqrt\pi\:((k-1)+\frac12)((k-2)+\frac12)\dots (1+\frac12) \frac12 }=\\ =2k \dfrac{1}{(1+\frac{1}{2(k-1)})(1+\frac{1}{2(k-2)})\dots (1+\frac{1}{2})}< 2k \dfrac{1}{(1+\frac{1}{2(k-1)})^{k-1}}=2k(1-\dfrac{1}{2k-1})^{k-1}<\\ <2k(1-\dfrac{1}{2k})^{k-1}<\dfrac{2k}{\sqrt e}(1-\dfrac1{2k})^{-1}<\dfrac{2k}{\sqrt e}\dfrac{6}{5}<\dfrac{\pi}{2}(k+1) $$ for $k>2$