Let $E_1$ and $E_2$ be two Elliptic curve defined over $\mathbb Q$ . Let $p$ be an fixed given odd prime of $\mathbb Q$ at which both the curves have good ordinary reduction. Moreover padic $L$function of $E_1$ and $E_2$ are same . Does it mean that the complex $L$function of $E_1$ and $E_2$ are also same ? Is there some sufficient criteria on padic $L$functions such that such that the $L$ function is determined?

Hmmm. I am not so sure that the answer is "No". In fact I would rather bet on "Yes". Of course, I totally agree that the characteristic ideal, i.e. the ideal generated by the $p$adic $L$function in $\Lambda$, is not enough to determine the elliptic curve. In particular there are plenty of curves for which the $p$adic $L$function is a unit in $\Lambda^{\times}$. The $p$adic $L$function can be viewed as a measure $\mu$ on the Galois group of $F_\infty/\mathbb{Q}$. It is build up from modular symbols of the form $\bigl[\frac{a}{p^k}\bigr]$ as $a$ and $k$ varies over all positive integers. Knowing the measure $\mu$ it is easy to extract the unit root $\alpha$ of the Frobenius at $p$ and hence the value of $a_p$. Then it is not difficult to see from the definition of $\mu$ that one can compute all the modular symbols $\bigl[\frac{a}{p^k}\bigr]$. It is true that these values do not seem to carry the value of $a_{\ell}$ for primes $\ell\neq p$ with them needed to reconstruct the complex $L$function; we would need modular symbols with $\ell$ in the denominator and I can not see immediately how to get them from $\mu$. Nevertheless, there are plenty of values of $a$ and $k$. And it would be a big surprise to me if there very by chance two elliptic curves such that all the values of the modular symbols $\bigl[\frac{a}{p^k}\bigr]$ would be equal. But I have no clue of how to prove this intuition. So I ran through some examples. Let $F_{\infty}$ be the cyclotomic $\mathbb{Z}_p$extension of $\mathbb{Q}$. I picked a few elliptic curves of small conductor such that
Then I computed the $p$adic $L$functions $L_p(T)$ for these with $T$ corresponding to $1+p$ under the cyclotomic character Gal$(F_{\infty}/\mathbb{Q})$. By what I have imposed the leading term will be equal. For each $p^n$th root of unity $\zeta$, the value of $L_p(\zeta1)$ is, up to a power of $\alpha$ which is the same for all my curves because I fixed $a_p$, equal to the order of the TateShafarevich group at the $n$th level; at least if one believes the ($p$adic version of the) Birch and SwinnertonDyer conjecture. From what I imposed, it is clear that the $p$primary part will be trivial, but there may be different primes appearing in Sha for various curves. So there is no reason to believe that it would be easy to find two curves that have the same $p$adic $L$function in these family that I have chosen. Here are some examples with $p=5$ and $a_5=1$. 139a1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (1 + 4 \cdot 5 + O(5^2)) \cdot T + (3 + 5 + O(5^2)) \cdot T^2 + (3 + 2 \cdot 5 + O(5^2)) \cdot T^3 + (1 + 5 + O(5^2)) \cdot T^4 + O(T^5)$ 141e1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (4 + 3 \cdot 5 + O(5^2)) \cdot T + (3 \cdot 5 + O(5^2)) \cdot T^2 + (5 + O(5^2)) \cdot T^3 + (2 + 4 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5) $ 346a1 $ 4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (2 + 5 + O(5^2)) \cdot T + (4 \cdot 5 + O(5^2)) \cdot T^2 + O(5^2) \cdot T^3 + (1 + 2 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5)$ 906i1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (3 + O(5^2)) \cdot T + (2 + 5 + O(5^2)) \cdot T^2 + (3 + 5 + O(5^2)) \cdot T^3 + (3 + 2 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5)$ Finally a word why I believe the answer should be Yes. A big dream in the direction of BSD is to hope that there is a link between the $p$adic and the complex $L$function. Of course, we should believe that the order of vanishing at $s=1$ should be equal for instance (and this is only known when it is at most a simple zero). I would hope that such a link will be bijective. But that is a mere intuition and hence possibly wrong. But it also explains that the answer to this question might well be very difficult. 


Yes. See the Invent. Math. paper by Luo and Ramakrishnan titled "Determination of modular forms by twists of critical Lvalues". They mention this result on the third page. 


This is not a definitive answer to the question. It's goal is simply to establish enough background and perspective on the question to explain why I think that the answer is likely to be ``No''. The $p$adic $L$function does not interpolate infinitely many special values of the $L$function of $E$. Rather, the $p$adic $L$function (and this is a feature of $p$adic $L$functions generally) interpolates critical values (in the sense of Deligne) of twisted $L$functions of $E$. For an elliptic curve, the only critical value is $s = 1$; it is because one allows twisting that one ends up with a function, and not just a number. A little more precisely: the $p$adic $L$function of an elliptic curve $E$ over $\mathbb Q$ interpolates the twisted central values, i.e. the values $L(f\_{\chi},1),$ where $f$ is the weight two modular form attached to $E$ by the modularity theorem for elliptic curves, $\chi$ denotes a character of $p$power conductor, and $f\_{\chi}$ denotes the twist of $f$ by $\chi$. (There are some extra factors to do with interpolation and so on, which I will ignore here.) More precisely again, supposing that $E$ has good ordinary reduction at $p$ (and perhaps some other technical conditions which I'll suppress) the $p$adic $L$function is an element of the completed group ring ${\mathbb Z}\_p[[{\mathbb Z}\_p^{\times}]]$, whose specialization under a finite order character $\chi:{\mathbb Z}\_p^{\times} \rightarrow \overline{\mathbb Q}^{\times}$ is (essentially) the algebraic part of the classical $L$value $L(f\_{\chi},1)$. Now the complex $L$function of $E$ determines (and is determined by) the isogeny class of $E$. The reason for this is that a consideration of the Euler product allows one to determine the traces of all Frobenius elements on the Tate modules of $E$, and hence by Faltings' theorem (the Tate conjecture) determine $E$ up to isogeny. On the other hand, the $p$adic $L$function does not have a corresponding Euler product, and so it is not clear (to me, at least) that it determines the $p$adic Tate module of $E$ (and hence $E$, up to isogeny, or equivalently, the complex $L$function of $E$). One complication (at least at a psychological level) in thinking about this question is that in Iwasawa theory one often just thinks about the ideal in ${\mathbb Z}\_p[[{\mathbb Z}\_p^{\times}]]$ that is generated by the $p$adic $L$function, since it is this ideal which is (according to the main conjecture) supposed to relate to the Selmer group of $E$ over the $p$adic cyclotomic tower. And this ideal surely won't determine $E$ up to isogeny; it is much too coarse a piece of information (e.g. it could just be the unit ideal in many cases, say if $E$ has no points over ${\mathbb Q}(\zeta\_p)$ and the $p$torsion part of Sha$(E)$ over this field is trivial (and maybe some condition on Tamagawa numbers).) The $p$adic $L$function carries more information than this ideal, of course; it really does know about all those special values. But it's not clear to me exactly what to do with this information. So while I'm pretty sure that the answer to your question is ``No'', I'm not sure how to actually prove it either way. 


Yes, it is true that the $p$adic $L$function of an elliptic curve $E$ over $\mathbf{Q}$ determines the isogeny class of $E$. As mentioned in the LuoRamakrishnan article pointed out by Idoneal, this follows from analytical results of Rohrlich. I thought I would sketch the argument of Rohrlich because it's interesting. So, let us assume $L_p(E_1,\cdot)=L_p(E_2,\cdot)$ for two elliptic curves $E_1,E_2$ with good ordinary reduction at $p$. By the interpolation property of the $p$adic $L$function, this means that \begin{equation} \frac{L(E_1,\chi,1)}{\Omega_{E_1}^\pm} = \frac{L(E_2,\chi,1)}{\Omega_{E_2}^\pm} \end{equation} for all Dirichlet characters $\chi$ of $p$power conductor. Here $\pm = \chi(1)$ and $\Omega^+$ (resp. $\Omega^{}$) is the real (resp. imaginary) period of a Néron differential. Now there is this nice analytical result by Rohrlich (On $L$functions of elliptic curves and cyclotomic towers, Invent. Mat. 75, 1984) : if $\chi$ is a Dirichlet character of conductor $p^m$, then the average value of $L(E,\chi^{\sigma},1)$, where $\chi^\sigma$ runs through the conjugate characters of $\chi$, tends to $1$ as $m \to +\infty$. This is proved by writing $L(E,\chi,1)$ as a quickly converging series and using clever analytical arguments. From this we deduce that $\Omega_{E_1}^{\pm}=\Omega_{E_2}^{\pm}$. This implies that there is an isogeny $\varphi : E_1 \to E_2$ (which is a priori defined over $\mathbf{C}$) such that $\varphi^* \omega_{E_2}$ is a nonzero rational multiple of $\omega_{E_1}$. By taking the sums of the conjugates of $\varphi$, we deduce that $E_1$ and $E_2$ are isogenous over $\mathbf{Q}$ (and thus their complex $L$functions coincide). Remark : we cannot always deduce that $E_1$ and $E_2$ are isomorphic over $\mathbf{Q}$. For example $E_1=15a1$ and $E_2=15a4$ are $2$isogenous and have the same Néron periods $\Omega_E^{\pm}$, so they have the same $p$adic $L$function, but they are not isomorphic. To me this rather indicates that the definition of $p$adic $L$function is not the right one (as pointed out by Olivier in his comment). If we want a theory of $p$adic $L$functions which looks like as much as possible as the theory of complex $L$functions, then the $p$adic $L$function should be invariant under isogeny. 

