Here are some useful hints, given that I think this is a homework question you're asking about. Use d'Alembert's formula for the solution for $u(x,t)$, which gives
$$ u(x,t) = \frac{1}{2}\left[g(x+t) + g(x-t)\right] + \frac{1}{2} \int_{x-t}^{x+t} h(y) \, dy$$
Now calculate $u_t$ and $u_x$ from this, plug those into your expressions for $k(t)$ and $p(t)$, and try calculating $k(t) - p(t)$ for large time $t$. Remember that you're given that $g,h$ are compactly supported functions, so somewhere down the line you're going to be using this information.
(a) We define$$
e(t)\equiv k(t)+p(t)=\frac12\int_{-\infty}^\infty \left( u_t^2+u_x^2\right)\,dx.
$$
Since $g,h$ have compact supports, we have that\begin{align*}
\frac{d}{dt}e(t)&=\frac12\int_{-\infty}^\infty 2u_tu_{tt}+2u_xu_{xt}\,dx\\
&=\int_{-\infty}^\infty u_tu_{tt}\,dx-\int_{-\infty}^\infty u_{xx}u_t\,dx\\
&=\int_{-\infty}^\infty u_t(u_{tt}-u_{xx})\,dx=0.
\end{align*}
Hence, $e(t)\equiv e(0)$ and so $k(t)+p(t)$ is constant in $t$.
(b) By d'Alembert's formula, we have$$
u(x,t)\frac12 (g(x+t)+g(x-t))+\frac12 \int_{x-t}^{x+t} h(y)\,dy.
$$
Thus\begin{align*}
u_t&=\frac12 (g'(x+t)-g'(x-t))+\frac12(h(x+t)+h(x-t)),\\
u_x&=\frac12 (g'(x+t)+g'(x-t))+\frac12(h(x+t)-h(x-t))
\end{align*}
We assume that there exists a positive constant $M$ so that $[-M,M]\supseteq supp(g')$ and $[-M,M]\supseteq supp(h)$. Note that for a fixed $t>M$,$$
-M\leq x-t\leq M\Leftrightarrow 0<t-M\leq x\leq t+M$$and $$-M\leq x+t \leq M\Leftrightarrow -t-M\leq x\leq -t+M<0.$$
Thus, when $t>M$,
$\,\,\,\,\,$(i) $0<t-M\leq x\leq t+M$. Then,$$
h(x+t)=g(x+t)=0
$$and so$$
u_t^2=\frac14 g'(x-t)^2+\frac14 h(x-t)^2-\frac12 g'(x-t)h(x-t)=u_x^2.
$$
$\,\,\,\,\,$(ii) $-t-M\leq x\leq -t+M<0$. Then,$$
u_t^2=\frac14 g'(x+t)^2+\frac14 h(x+t)^2+\frac12 g'(x+t)h(x+t)=u_x^2.
$$
$\,\,\,\,\,$(iii) Otherwise,$$
g'(x+t)=g'(x-t)=h(x+t)=h(x-t)=0.
$$
Hence, combining all the cases, it follows that, when $t>M$, $k(t)=p(t)$.
Best Answer
Define $E_k(t) \equiv \frac{1}{2}\int_{\mathbb{R}}u_t^2(x,t){\rm d}x$ and $E_p(t) \equiv \frac{1}{2}\int_{\mathbb{R}}c^2u_x^2(x,t){\rm d}x$. Then taking the derivative and using integration by parts we get $$\frac{d}{dt}(E_k(t) + E_p(t)) = \int_{\mathbb{R}}u_t[u_{tt} - c^2u_{xx}]{\rm d}x = 0$$ so the total energy is conserved. To show that $E_p(t) = E_k(t)$ for large $t$ we need an expression for the solution. This is given by d'Alemberts formula
$$u(x,t) = \frac{\phi(x+ct) + \phi(x-ct)}{2} + \frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s){\rm d}s$$
from which you can compute $u_t$, $u_x$ and derive
$$E_k(t) - E_p(t) = \frac{1}{2}\int_{\mathbb{R}}[\psi(x+ct)+c\phi'(x+ct)][\psi(x-ct)-c\phi'(x-ct)]{\rm d}x$$
To make the algebra above simpler note that $u_t^2 - c^2u_x^2 = (u_t - cu_x)(u_t+cu_x)$. Now since $\phi$ and $\psi$ has compact support then there is a $M>0$ such that $\phi'(x) = \psi(x) = 0$ for all $|x| > M$. Now if $ct > \frac{M}{2}$ then either $|x-ct| > M$ or $|x+ct|>M$ for all $x$ so the integrand above is identical to zero and $E_p = E_k$ follows. Finally since we know that $E_k + E_p$ is constant it follows that $E_p$ and $E_k$ has to be constant for large $t$.