Note that your description $kd(x,y)\leq p(x,y)\leq td(x,y)\tag{1}$ means that $p$ and $d$ are strongly equivalent metrics. For such metrics, it's fairly easy to show that we have the same convergent sequences.
Let $(x_n)$ be any sequence of points of $X$.
On the one hand, suppose that $p(x_n,x)\to0$ for some $x\in X$. Take any $\epsilon>0$. To show that $d(x_n,0)\to 0$, we must show that there exists some $N_1$ such that $d(x_n,x)<\epsilon$ for all $n\geq N_1$. Since $p(x_n,x)\to 0$, then there exists $N_1$ such that $p(x_n,x) for all $n\geq N_1$, and so by $(1)$, $d(x_n,x)=\frac1k\cdot kd(x_n,x)\leq\frac1k\cdot p(x_n,n)<\frac1k\cdot k\epsilon=\epsilon$ for all $n\geq N_1$.
On the other hand, suppose $d(x_n,x)\to 0$ for some $x\in X$. Taking $\epsilon>0$ and finding $N_2$ such that $d(x_n,x)<\frac{\epsilon}t$ for all $n\geq N_2$, we again use $(1)$ to see that $p(x_n,x)<\epsilon$ for $n\geq N_2$.
Thus, the metrics share the same convergent sequences.
It isn't actually necessary to use the constants $k,t$--in fact, it's even okay if there are no positive constants $k,t$ satisfying $(1)$ for all $x,y\in X$.
We say that metrics are equivalent if they produce the same open sets. Put rigorously, we mean the following:
(i) $\forall x,y,z\in X$ $\forall R>0$, if $d(x,y), then there exists $r>0$ such that $d(x,z) whenever $p(y,z). (Every point in an open $d$-ball lies in an open $p$-ball contained in the $d$-ball.)
(ii) [Same thing as (i), but swapping $p$ and $d$ in each instance.]
Strongly equivalent metrics are equivalent, but the converse needn't hold.
Suppose that $p$ and $d$ are equivalent, and let $(x_n)$ is a sequence of points of $X$.
On the one hand, suppose $p(x,x_n)\to 0$, and take $\epsilon>0$. Since $p$ and $d$ are equivalent, then there is some $r>0$ such that $d(x,x_n)<\epsilon$ whenever $p(x,x_n). Since $p(x,x_n)\to 0$, there is some $N$ such that $p(x,x_n) for all $n\geq N$, so $d(x,x_n)<\epsilon$ for all $n\geq N$. Thus, $d(x,x_n)\to 0$.
A similar approach will let you show that $d(x,x_n)\to0$ implies $p(x,x_n)\to 0$.