Calculating the extraordinary cohomology of $\mathbb{C}P^n$ Let $E$ be a ring spectrum with an orientation. Now I want to calculate  $E^*(\mathbb{C}P^n)$.
The definition of orientation I am using is: There is an element $x \in E^*(\mathbb{C}P^{\infty})$ such that it's restriction to $\mathbb{C}P^1$ is a generator over $\pi_*(E)$.
To calculate this I want to use the Atiyah-Hirzebruch spectral sequence. But I am unable to figure out why all the differentials vanish. I believe one has to some how use the fact that $E$ has an orientation but I am unable to figure out how. Thank you.
 A: (I'll use reduced cohomology for this answer for mild convenience.) $\require{AMScd}$For any ring spectrum $E$, the Atiyah-Hirzebruch spectral sequence for (the reduced groups) $E^*(\Bbb{CP}^1)$ degenerates at the $E^2$ page for degree reasons; this can be seen as coming from the (definitional in spectra) isomorphism $\pi_*(\Sigma^2 E) = \pi_{*-2}(E)$. Let the unit be $1 \in E^2(\Bbb{CP}^1) = \pi_0(E)$. 
There are now maps of spectral sequences associated to the inclusions $\Bbb{CP}^1 \hookrightarrow \Bbb{CP}^n \hookrightarrow \Bbb{CP}^\infty$. Now assume the ring spectrum is complex oriented. Let $x \in E^2(\Bbb{CP}^\infty)$ restrict to $1 \in E^2(\Bbb{CP}^1).$ We have a commutative diagram
\begin{CD}\pi_*(E) @>>> AH_2^{*,2}(\Bbb{CP}^\infty) @>>> AH_2^{*,2}(\Bbb{CP}^1) \\
@VV=V @VVV @VVV \\
\pi_*(E) @>\sigma \mapsto x \sigma>> E^{2+*}(\Bbb{CP}^\infty) @>>> E^{2+*}(\Bbb{CP}^1)
\end{CD}
where the top map identifies $\pi_*(E)$ with the bottom nonzero row in the Atiyah-Hirzebruch spectral sequence. Because the composite map given in the bottom row is an isomorphism, that entire row of the Atiyah-Hirzebruch spectral sequence for $\Bbb{CP}^\infty$ must survive to $AH_\infty$. But it is easy to check using the multiplicative structure of the spectral sequence that if any differential is nonzero, the transgression (whose input lies on $AH_n^{0,n+2}$ and output lies on $AH_n^{n+1,2}$) is nonzero; but if this were true, that would contradict the fact that the entirety of the bottom row survives to the $AH^\infty$. So the spectral sequence for $\Bbb{CP}^\infty$ degenerates on the second page. You see the same for $\Bbb{CP}^n$ by investigating the map of spectral sequences associated to the inclusion $\Bbb{CP}^n \hookrightarrow \Bbb{CP}^\infty$. 
So the unreduced cohomology $E^*(\Bbb{CP}^\infty) = \pi_*(E)[x]$, where $|x| = 2$ (I guess this is usually written instead as power series). $E^*(\Bbb{CP}^n)$ is, accordingly, $\pi_*(E)[x]/(x^{n+1})$.

You might see Lurie's brief notes on complex oriented cohomology theories.

Stable cohomotopy is an explicit example of a non-complex-oriented cohomology theory for which the spectral sequence does not degenerate. The Segal conjecture for compact Lie groups (here) implies that after $p$-completion, $\pi_s^0(\Bbb{CP}^\infty_p) = \Bbb Z_p$, the $p$-adics (since the Burnside ring of $S^1$ is just $\Bbb Z$). If the spectral sequence degenerated, then the 0th stable cohomotopy of $\Bbb{CP}^\infty_p$ would have a filtration such that the subquotiens are $\pi_s^{2k}(S)$. But this is not possible, because $\pi_s^{2k}(S)$ can have summands like $(\Bbb Z/p)^2$, and the only groups that can arise as subquotients of $\Bbb Z_p$ are $\Bbb Z_p$ itself or $\Bbb Z/p^k$.
A: Your condition should be that the orientation is an element of the reduced $E$ cohomology of $CP^\infty$, restricting to a generator of $\pi_0(E)$.
Look at the reduced AHSS. The elements with bidegree $(2,0)$ are perm cycles in the SS for $CP^1$ for degree reasons. The information about the orientation tells you they have to be perm cycles in the SS for $CP^\infty$ too and, in particular, in the SS for $CP^n$.
Since we have a splitting $* \to CP^n \to *$ the reduced AHSS sits inside the unreduced SS, and in the unreduced SS,
elements with $s=0$ are perm cycles.
Now multiply.
