Factoring Anderson's RSA backdoor In 1993, Anderson [1] proposed a backdoor to the RSA key generation algorithm. This backdoor requires that a backdoor key (prime) $A$ be implanted within the key generation portion of the RSA algorithm.
Instead of the usual way, primes $P$ and $Q$ are generated using the following algorithm:
First define a backdoor prime $A$ and two smaller random primes $P'$ and $Q'$.
Let $k=1$
$ \text{If} \ \ isprime(A\cdot k + P'):\\
\quad P = A\cdot k + P' \\
\text{else}: k = k+1$
The analog is performed for $Q$ using $Q'$.
This algorithm was also described here Is there any more information on this RSA backdoor?
This backdoor allows to calculate $N′= N \mod A$ and then factor $N′$ into $P′$ and $Q′$. Still $N'$ needs to be factored, but this is now a much easier problem as $N'$ is only about a fourth in size of $N$.
Please note that in my algorithm above I used $k=1$, the original implementation by Anderson suggests starting value $k=P'$ and iteratively increasing $k$ by one until $P$ is prime. In my algorithm I start with $k=1$.
My questions are:

does starting the iteration from $k=1$ instead of $k=P'$ make a difference?
in the way $N'$ is generated, what is the best way to factor $N'$ given the information how it is generated? Is there a certain factoring algorithm that makes factoring $N'$ very easy?

[1] Ross Anderson. Practical RSA Trapdoor. Electronic Letters. 29(11): 995,
1993.
does starting the iteration from $k=1$ instead of $k=P'$ make a difference?


If you start iteration from $k=P'$ then you get;
$$P = A\cdot (P'+i) + P'$$
where $ i = 0,1,2,\ldots$. Take modulo $A$
$$P  = A\cdot (P'+i) + P' \pmod A$$
$$P  =  P' \pmod A$$
Therefore it will still work to reveal the $P'$


in the way $N'$ is generated, what is the best way to factor $N'$ given the information how it is generated? Is there a certain factoring algorithm that makes factoring $N'$ very easy?


The current factoring record in the open literature is 829-bit, however, the recommended key size is at least 2048, i.e each factor has 1024-bits. Therefore the size of $A$ must be around $1024$-bit. Once you set $A$, then you can seek $P = A\,k_1 + P'$ where $P$ is prime.
There is nothing that prevents one to generate small primes for $P'$  and $Q'$. The $P'$ and $Q'$ doesn't add much to $P = A\,k_1 + P'$ since they are added. The $A$ and $k_1$ is important.
Therefore you can choose $P'$ and $Q'$ as below as $829$-bits.
You can use the CADO-NFS to factor the $P'\cdot Q'$. Here some notes about it, too.