Applying Chinese RemainderTheorem and Paillier Homomorphic encryption
A working method for CRT-aided decryption in Pailler goes:


Evaluate $x=c^\lambda\bmod n^2$ by the Chinese Remainder theorem, that is


$x_p\gets c^\lambda\bmod p^2$
$x_q\gets c^\lambda\bmod q^2$
$x\gets\left(q^{-2}(x_p-x_q)\bmod p^2\right)q^2+x_q$

Then evaluate $m\displaystyle\gets\left\lfloor\frac{x-1}n\right\rfloor\,\mu\bmod n$.


The savings are by a factor of at most 2, rather than at most 4 for RSA, because the exponent $\lambda$ remains the same size, when its $d_p$ and $d_q$ have their size halved in RSA.



The question's method is broken where it goes from the following line to the next:

$\begin{align}
m&amp;=L\left(c^{(p-1)(q-1)}\bmod n^2\right)\cdot\left((p-1)(q-1)^{−1}\bmod n\right)\bmod n\\
m_p&amp;=L_p\Big(\left(c^{p-1}\bmod p^2\right)\cdot\big((p-1)^{−1}\bmod p\big)\Big)\bmod p
\end{align}$

Problems are:


Function $L_p$ is not stated. I work under the assumption that $L_p$ is to $p$ what $L$ is to $n$, that is $L_p(x)\displaystyle=\left\lfloor\frac{x-1}p\right\rfloor$. Also I assume $m_p=m\bmod p$.
The term $(q-1)^{-1}$ of the first equation mysteriously vanishes from the second.
Something on the tune of $L(x)\equiv L_p(x)\bmod p$ seems to be assumed  without justification.
The term $(p-1)^{-1}\bmod p$ (which BTW is always $p-1$) has mysteriously entered the input of $L_p$ when $(p-1)^{-1}$ was at the output of $L$.


Numerical experiments disprove the second equation (under the assumptions made in 1). If there is a way to repair it, I want to know!