I'm looking for the answer to following question. Do exist different knots in $RP^3$ which have equivalent liftings in $S^3$ under covering $p:S^3\rightarrow RP^3$?

2$\begingroup$ Could you give us some context for your interest in the question? My concern is it sounds a lot like a homework problem. $\endgroup$ – Ryan Budney Mar 13 '13 at 17:41
If either knot is hyperbolic, this is not possible.
First, consider the nonnull homologous case. Let $K_1,K_2\subset \mathbb{RP}^3$ be two knots, such that their preimages $K_1',K_2'\subset S^3$ are isotopic to the hyperbolic knot $K$ (Remark: if $K_1$ is hyperbolic, then so is $K$ and therefore $K_2$ by the geometrization theorem). The covering translation induces involutions $\iota_{1,2}:S^3 \to S^3$ such that $\iota_{i}(K)=K$. Restricting to the hyperbolic space $S^3\backslash K$, we get involutions $\iota_i:S^3\backslash K \to S^3\backslash K$, $i=1,2$, which are isotopic to (fixedpoint free) hyperbolic isometries. The hyperbolic isometry restricts to the torus $T=\partial\mathcal{N}(K)$ as an isometry (e.g. taking a horotorus representative). Each isometry is determined by its action on $T$. There are 3 possible fixedpoint free involutions of a torus. Let $\mu, \lambda\subset T$ be representatives of the meridian and longitude. The involutions rotate around $\mu$ (preserving $\mu$), or rotate around $\lambda$, or rotate around $\mu\lambda$. The rotation around $\mu$ is not possible, since this would extend to an involution of $S^3$ with fixedpoint set $K$, which is impossible by the Smith conjecture. Thus, $\iota_1$ and $\iota_2$ must induce the other two involutions. But then $\iota_1\circ\iota_2$ must be the forbidden involution, a contradiction. So there is a unique involution preserving $S^3\backslash K$, and therefore at most one isotopy class of knot in $\mathbb{RP}^3$.
In the null homologous case, there are knots $K_1,K_2\subset \mathbb{RP}^3$ such that the preimage is isotopic to a twocomponent link $L$. There are two fixedpoint free involutions of $S^3$ preserving $L$ and exchanging its components, and isotopic to isometries of the hyperbolic metric on $S^3\backslash L$ which exchanges the two cusps. Thus, they generate a dihedral group in the isometries of the hyperbolic knot complement, which must be finite order. Since both isometries preserve the meridians of $L$, this may be extended to a dihedral group action on $S^3$, generated by two fixed point free involutions. However, any such (smooth) action is conjugate to a group of isometries of $S^3$ by the orbifold theorem, which gives a contradiction, since the only fixedpoint free involution is the antipodal map.
I think the general case should follow in a similar fashion, but one would have to consider how the involutions permute the JSJ decomposition.

1$\begingroup$ Given Ryan's comment, I'm probably missing some simpler argument, but I've been a bit sleepdeprived lately. $\endgroup$ – Ian Agol Mar 13 '13 at 19:05

$\begingroup$ @Agol Are you assuming the knots $K_1$, $K_2$ are nonnull homologous in $\mathbb{RP^3}$? It seems if the knots are null homologous, $p^{1}(K_i)$ could be a two component link. $\endgroup$ – Neil Hoffman Mar 14 '13 at 13:41

$\begingroup$ @Neil: yes, I was only considering that case, although clearly one should consider the null homologous case too. $\endgroup$ – Ian Agol Mar 14 '13 at 15:57

$\begingroup$ Ok, I've added some discussion of the null homologous case. In fact, these could probably be treated in a uniform manner, since the Smith conjecture follows from the orbifold theorem. $\endgroup$ – Ian Agol Mar 14 '13 at 19:26