Multiplying the second equation by $y$ and using the first equation we have
$$y\dot{y}=-2xy-p(t)y^4=-x\dot{x}-p(t)y^4$$
i.e. as correctly pointed in the comments by @Evgeny
$$\frac{d}{dt}[x^2+y^2]=-2p(t)y^4\leq -4 y^4$$
Thus $x,y$ are bounded and the Lyapunov-like function $V=x^2+y^2$ is decreasing and lower bounded (from zero). Hence, $V$ converges to some constant $V_{\infty}\geq 0$. If we integrate the above inequality over $[0,\infty)$ we obtain
$$V_{\infty}-V(0)\leq -4\int_0^{\infty}{y^4(s)ds}$$
which yields
$$\int_0^{\infty}{y^4(s)ds}\leq \frac{1}{4}V(0)$$
If we assume a bounded $p(t)$ then the boundedness of $x,y$ and the state equations result in the boundedness of $\dot{x},\dot{y}$.
A continuous differentiable function with bounded derivative is uniformly continuous.
You can use now Barbalat's lemma to prove convergence. Barbalat's lemma states that if
i) $\int_0^{\infty}{\phi(t)dt}$ exists and is finite and
ii) $\phi(\cdot)$ is uniformly continuous
then $\lim_{t\rightarrow\infty}\phi(t)= 0$.
From Barbalat lemma we then have that $\lim_{t\rightarrow\infty}y(t)=0$ and since $V$ converges there also exists some $x^*$ (with $V_{\infty}={x^*}^2$) such that $\lim_{t\rightarrow\infty}x(t)=x^*$.
Edit to remove the upper bounded $p(t)$ restriction: I will prove now a variation of Barbalat lemma that does not need uniform continuity of $\phi(\cdot)$ but only an upper bounded derivative.
Barbalat lemma variation: Let $\phi:\mathbb{R}\rightarrow\mathbb{R}_+$ continuous differentiable nonnegative function. If
i) $\int_0^{\infty}{\phi(t)dt}$ exists and is finite and
ii) $\dot{\phi}(\cdot)$ is upper bounded
then $\lim_{t\rightarrow\infty}\phi(t)= 0$.
Proof: The proof uses a contradiction argument. Assume the opposite, then there exists a constant $k_1>0$ and a sequence of times $\{T_i\}$ with $\lim_{i\rightarrow\infty}T_i=\infty$ such that
$$\phi(T_i)\geq k_1$$
For $t\leq T_i$ we have from the mean value theorem that
$$\phi(t)=\phi(T_i)-\dot{\phi}(\theta t +(1-\theta)T_i)(T_i-t)$$
for some $\theta\in(0,1)$. Since $\dot{\phi}$ is upper bounded there exists some $c$ such that $\dot{\phi}(t)\leq c$ for all $t\geq 0$ and
$$\phi(t)\geq \phi(T_i)-c(T_i-t)\qquad \forall t\leq T_i$$
From the above relationship we have that
$$\phi(t)\geq \frac{k_1}{2}\qquad \forall t\in\left[T_i-\frac{k_1}{2|c|},T_i\right]$$
Therefore
$$\int_{T_i-\frac{k_1}{2|c|}}^{T_i}{\phi(t)dt}\geq \frac{k_1^2}{4|c|} $$
Thus $\int_0^{t}{\phi(\tau)d\tau}$ cannot converge to a finite limit as $t\rightarrow \infty$ which is the desired contradiction.
In our example $\phi(t):=y^4$ and its derivative is upper bounded since
$$\dot{\phi}(t)=4y^3\dot{y}=-8xy^3-4p(t)y^6\leq -8xy^3$$
and $x,y$ are bounded. The proposed variation of Barbalat lemma yields now the desired $\lim_{t\rightarrow\infty}y(t)=0$. This completes the proof and indeed there is no need for an upper bound of $p(t)$.
Best Answer
Arnold’s book “Ordinary Differential Equations” is absolutely fantastic and contains many of the topics you are looking for.