Why doesn't momentum eigenstates and wavevector eigenstates have the same form in position representation? Let $|k'\rangle$ and $|p'\rangle$ be eigenstates of the operator $k$ and $p$ with corresponding eigenvalue $k'$ and $p'$, where $k$ and $p$ represent wavevector and momentum  operators.
Since $k$ and $p$ commute, then they have the common set of simultaneous eigenvectors. It's easy to verify that both {$|k'\rangle$} and {$|p'\rangle$ are eigenstates of either operator $p$ or $k$. So we can write both of the sets as :{$|p',k'\rangle$}.
However there must be a logic flaw here, if this is true, then this means that $\langle x'|k'\rangle$ is the same as $\langle x'|p'\rangle$ in position representation because they can be both written as  $$\langle x'|p',k'\rangle$$
But  $\langle x'|k'\rangle$  is:

$\langle x'|p'\rangle$ is:

Why they are differ by a $\sqrt{\hbar}$ in the coefficient? (note the picture for  $\langle x'|k'\rangle$ is 3D case, but for $\langle x'|p'\rangle$ it's 1D case. For simplicity, let's talk about 1D case.
PS: I can derive $\langle x'|k'\rangle$ and $\langle x'|p'\rangle$, it is easy to get the normalization factors, but I what I want to ask is why the logic I used above is wrong?
 A: In 3 dimensions (with $\hbar = \hbar$), it is the case that 
$$\langle \mathbf{r} | \mathbf{p} \rangle = \frac{1}{(\sqrt{2\pi \hbar})^3} e^{-i \mathbf{p} \cdot \mathbf{r}/\hbar} $$
Next recall that $\langle \mathbf{r} | \mathbf{p} \rangle^* = \langle \mathbf{p} | \mathbf{r} \rangle$, and so 
$$\langle \mathbf{p} | \mathbf{r}\rangle = \frac{1}{(\sqrt{2\pi \hbar})^3} e^{i \mathbf{p} \cdot \mathbf{r}/\hbar}. $$

Derivation: I will do it for the 1D case. 
$$\langle x| \hat{p} | p \rangle = p \langle x | p \rangle \tag{1}$$
Next, the definition of the momentum operator as the generator of translations is given by  
$$ {\hat {p}} =i\hbar \lim _{\delta x \rightarrow 0}{\frac {{\hat {T}}(\delta x\mathbf {\hat {x}} )-{\hat {\mathbb {I} }}}{\delta x}} $$ 
where $\hat{T}$ is the translation operator. Therefore, since $\hat{p}$ is hermitian we can just apply it to $\langle x |$.
$$\langle x| \hat{p} = i\hbar \lim_{\delta x \to 0}\frac{ \langle x + \delta x| - \langle{x}|}{\delta x} $$
plugging this into (1) we have that 
$$i\hbar \lim_{\delta x \to 0}\frac{ \langle x + \delta x|p\rangle - \langle{x}|p\rangle}{\delta x}  = i\hbar \frac{d}{dx} \langle{x}| p\rangle = p \langle x | p \rangle $$
That is, we have the differential equation 
$$ 
i\hbar \frac{d}{dx} \langle{x}| p\rangle = p \langle x | p \rangle
$$
This is precisely solved by 
$$ \langle x | p \rangle = C e^{-i x p /\hbar} $$
for some constant $C$. 
A: That’s because the wavefunction is normalized to the Dirac delta, and when you define the normalization integral in the 2 cases:
$$\langle x|x’\rangle=\int_{-\infty}^\infty dp \psi_p^*(x)\psi_p(x’)=\delta(x-x’)$$
Or
$$\langle x|x’\rangle=\int_{-\infty}^\infty dk \psi_k^*(x)\psi_k(x’)= \delta(x-x’) $$
You have an additional factor of $\hbar$ coming from $dp=\hbar dk$ to get rid of;
The reason of the different constant is due to the fact that the identity operator can be built both with $\hat k$ and with $\hat p$ eigenstates:
$$\mathbb I=\int dp |p\rangle\langle p|$$
$$\mathbb I=\int dk |k\rangle\langle k|$$
So if we keep symmetry in the definition of the identity operator, we must have asymmetry in the definitions of $\langle p|x\rangle\equiv\psi_p(x)$ and $\langle k|x\rangle\equiv\psi_k(x) $;
