If $C$ is allowed to depend on $\lambda$, just take $C=2\lambda^p$.

If $C$ is not allowed to depend on $\lambda$, take any nonzero $f\in H_{0}^{1}(0,1)$ and let $\lambda\to\infty$. Then the left-hand side of your desired inequality will go to $\int_0^1|f(x)|^2\,dx>0$ whereas its right-hand side will go to $0$, so that your desired inequality will fail to hold.

$\newcommand\la\lambda\newcommand\al\alpha$If $C$ is allowed to depend on $\lambda$, just take $C=2\lambda^p$.

If $C$ is not allowed to depend on $\lambda$, take any nonzero $f\in H_{0}^{1}(0,1)$ and let $\lambda\to\infty$. Then the left-hand side of your desired inequality will go to $\int_0^1|f(x)|^2\,dx>0$ whereas its right-hand side will go to $0$, so that your desired inequality will fail to hold.

Let $f(x)=x(1-x)$ and let $\la\to\infty$. Then $$\int_{\la^{-k}}^1|f(x)|^2\,dx\to\frac1{30}$$ and $$\int_{\la^{-k}}^1 x^\al|f'(x)|^2\,dx\to h(\al):=\frac{\alpha ^2+\alpha +2}{\alpha ^3+6 \alpha ^2+11 \alpha +6},$$ so that the constant factor $L$ in the inequality $$\int_{\la^{-k}}^1|f(x)|^2\,dx \le L\,\int_{\la^{-k}}^1 x^\al|f'(x)|^2\,dx$$ must be $\ge\dfrac1{30h(\al)}$. So, $L$ cannot go to $0$ as $\la\to\infty$.