EDIT: This is just an approach, it doesn't actually work. I don't know if it's possible to make it correct.
Let $q = \frac{1}{1 - \frac{1}{p}}$ so $\frac{1}{p} + \frac{1}{q} = 1$.
By Hölder's inequality we have
$$
\int_{0}^{x} |f(\lambda)| d\lambda \le (\int_{0}^{x} |f(\lambda)|^{p} d\lambda)^{\frac{1}{p}} (\int_{0}^{x} |1|^{q} d\lambda)^{\frac{1}{q}} \le ||f||_{p} x^{\frac{1}{q}}
$$
Hence
$$
|\frac{1}{x} \int_{0}^{x} f(\lambda) d\lambda|^p \le |\frac{1}{x} \int_{0}^{x} |f(\lambda)| d\lambda|^p \le ||f||^{p}_{p} x^{\frac{1}{q}-1}
$$
Now norm of $T(f)$ has the following form:
$$
||T(f)||_{p} = (\int _{0}^{\infty} |T(f)(x)|^{p}dx)^{1/p} = (\int _{0}^{\infty}|\frac{1}{x} \int _{0}^{x} f(\lambda) d\lambda|^pdx)^{1/p}
$$
Let $||f||^{p} = c$.
So
$$
||T(f)||_{p} \le (\int _{0}^{\infty} c^{p} x^{\frac{1}{q}-1} dx)^{1/p} = ||f||_p (\int _{0}^{\infty} x^{-1/p})^{p}
$$
Since $(\int _{0}^{\infty} x^{-1/p})^{p}$ is some constant we have $||T(f)|| _{p} \le ||f|| _{p} const$, hence this operator is bounded.