Is there a fast way to solve $k = n \cdot g^a \mod P$? (get $a$ for unknown $n$)

Question

Would a factor besides the normal discrete logarithm problem increase or decrease the solving time?

$k = n \cdot g^a \mod P$

with given $k,g,P$ and the knowledge $P= 2 \cdot N \cdot f+1$, while $f$ can be a product out of other primes. The factor $n<P-1 \in \mathbb{N}$. The generator $g$ can generate a group with max size of $N$.
How can we solve this?

Harder than solving the normal: $k' = h^a \mod Q$, with h prime rooot of $Q$?

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edit: forgot to mention at least two equations need to be solved with same $n$
$k' = n \cdot g^b \mod P$
or one without
$k' = k \cdot g^c \mod P$

Answer 1

$k = n \cdot g^a \mod P$

How can we solve this?

It is trivial to find $(n, a)$ pairs that satisfy this relation; select an arbitrary $a$ and compute $n = k g^{-a} \bmod P$; that's a solution.

Now, that'll give you $ord(g)$ distinct solutions; if you have a specific solution in mind, you have no way to telling which one it is. However, depending on how the solution is used, it might not matter...