$E_0(\Bbb{Q}_p)=\{P\in E(\Bbb{Q}_p)\mid \tilde{P} \in \tilde{E_{ns}}(\Bbb{F}_p)\}$, here $E_{ns}$ denotes sets of $E(\Bbb{Q}_p)$ which reduces to singular points of $\tilde{E}(\Bbb{F}_p)$.
Thus $E_0(\Bbb{Q}_p)=E(\Bbb{Q}_p)$ implies $E$ has good reduction at $p$.
But this LMFDB's date
https://www.lmfdb.org/EllipticCurve/Q/18496/n/2
reads $E:y^2=x^3+17x$ has bad reduction at $p=2$ but it has Tamagawa number $c_2:=[E(\Bbb{Q}_p):E_0(\Bbb{Q}_p)]=1$.
Where did I go wrong ?
On
Your first assertion is false. If $E$ has good reduction then the Tamagawa number is $1$, but the converse is not true in general. In particular since $c_p = [E(\mathbb{Q}_p), E_0(\mathbb{Q}_p)]$ is the number of connected components of the identity in the Neron model, in cases $I_1$ and $II$ certainly the Tamgawa number will be $c_p = 1$ (as you observed). I guess this can also happen in other cases depending on the Galois action.
To expand a bit on the answer by @Mummytheturkey:
The point is that $E(\mathbb{Q}_p)$ does not necessarily surject onto $\overline{E}(\mathbb{F}_p)$ when $\overline{E}$ is singular – Hensel's lemma says that the smooth points all lift, but the non-smooth ones might not.
In your example, the reduction is $y^2 = x(x + 1)^2$ over $\mathbb{F}_2$ which has a singular point $(1 : 0 : 1) \in \mathbb{P}^1(\mathbb{F}_2)$. If this were in the image of the reduction map, we'd have to have a solution $(x, y)$ of $y^2 = x^3 + 17x$ in $\mathbb{Z}_2$ with $x = 1$ and $y = 0$ mod 2; but this has no solutions mod 4, and hence none in $\mathbb{Z}_2$.