How does Dirac form this conjugate imaginary equation? On page 30 of Dirac's book
$$\xi|P\rangle = a|P\rangle\tag{12}$$
He then says

Suppose we have a solution of (12) and we form the conjugate imaginary equation, which will read
$$\langle P|\xi = a\langle P|$$
in view of the reality of $\xi$ and $a$

To show this, I tried from (12)
$$\langle P|\xi|P\rangle = \langle P|a|P\rangle$$
$$\langle P|\overline\xi|P\rangle = \langle P|\overline a|P\rangle$$
And since $a$ and $\xi$ are both real
$$\langle P|\xi|P\rangle = \langle P|a|P\rangle$$
But I don't see how I can now remove the $|P\rangle$ on both sides to get the conjugate imaginary equation.
 A: The sort of trick involved in 

removing the $|P\rangle$ on both sides to get the conjugate imaginary equation
$$\langle P|\xi|P\rangle = \langle P|a|P\rangle \tag1 $$

is quite common but it is indeed nontrivial to grasp the first time. In essence, you leverage the fact that in an equation of the form 
$$
⟨\psi|\hat A|\phi⟩=⟨\psi|\hat B|\phi⟩\tag2
$$
you have two linear functionals $⟨\psi|\hat A$ and $⟨\psi|\hat B$ which give the same result for all states $|\phi⟩\in\mathcal H$. This is exactly the criterion for equality of functions:

$f=g$ if and only if $f(x)=g(x)$ for all $x$ (modulo domain considerations). 

Therefore, you can conclude from (2) that $⟨\psi|\hat A=⟨\psi|\hat B$, which effectively means that you can "cancel out" the $|\phi⟩$ as long as the equality is valid for all such states.
That said, you've gotten yourself in some trouble in (1), because $P$ appears on both sides. This can maybe be fixed from there, but your best bet is to take a few steps back. Instead of multiplying your original equation
$$
\xi|P⟩=a|P⟩ \tag {Dirac's (12)}
$$
on the left by $⟨P|$, take its inner product with an arbitrary state $|\phi⟩$:
$$
⟨\phi|\xi|P⟩=⟨\phi|a|P⟩
$$
and then take the conjugate to conclude, via hermicity of $\xi$ and reality of $a$, that 
$$⟨P|\xi|\phi⟩=⟨P|a|\phi⟩$$
for arbitrary $|\phi⟩$. From this, then you can conclude that 
$$⟨P|\xi=⟨P|a.$$
