Power spectral density of a PAM signal
Since $x(t)$ is a random process, its Fourier transform generally doesn't exist. In its original form, the process $x(t)$ doesn't even have a power spectrum in the conventional sense because it is cyclostationary (and not stationary). It can be made stationary by adding a random phase $\theta$ that is uniformly distributed on $[0,T]$ and independent of $a_n$:

$$x'(t)=x(t+\theta)=\sum_na_nh(t+\theta-nT)\tag{1}$$

This process is stationary and its power spectrum can be computed, which is usually done by first determining its autocorrelation function, and then applying the Fourier transform to the autocorrelation function. The corresponding derivation can be found in most textbooks on digital communications:

$$S_{x'}(f)=\frac{1}{T}|H(f)|^2S_a(f)\tag{2}$$

where $H(f)$ is the Fourier transform of the pulse $h(t)$, and $S_a(f)$ is the power spectrum of the discrete-time random process $a_n$.

If we can assume that the symbols $a_n$ are uncorrelated, the power spectrum $S_a(f)$ is constant, $S_a(f)=P_a$ (where $P_a$ is the power of $a_n$), and $(2)$ becomes

$$S_{x'}(f)=\frac{1}{T}|H(f)|^2P_a\tag{3}$$

Finally, the power (not energy) of $x'(t)$ is the integral over its power spectrum (note that $h(t)$ is band-limited):

$$P_{x'}=\int_{-\infty}^{\infty}S_{x'}(f)df=\frac{1}{T}P_a\int_{-\infty}^{\infty}|H(f)|^2df=\frac{1}{T}P_aE_h\tag{4}$$

where $E_h$ is the energy of the impulse $h(t)$.