What probability distribution describes getting at least $k$ successes in a row, in $n$ trials? I am trying to figure which discrete probability distribution is it, here is the description:
A event had 2 possibilites: success or failure. The probability for success is $p$ and for failure is $1-p$  as usual
What is the probability of get at least $k$ success in a row $n$ trials?
E.g., with $k=3$, $n=12$:
F=failure              S=success

SFFSSFFFSSFS=NOK (zero row of 3 success)

SFF(SSS)FFSSFS=OK (one row of 3 success)

SSSSFFFFFFFF=OK (2 rows of 3 success, (SSS)SF... and S(SSS)F...)

SSSSSSSSSSSS=OK  (multiples 3 success in a row)

SFSFSFSFSFSF=NOK (zero row of 3 success)

SSFSSFSSFSSF=NOK (zero row of 3 success)

 A: You can compute your probabilities with a Markov chain.
In you example, consider a finite automaton with four states: $S_0,S_1,S_2,S_3$. Every state encodes the number of consecutive successes achieved so far, so when starting you go with probability $p$ in $S_1$ and with probability $(1-p)$ in $S_0$. If you are in $S_0$, you go with probability $p$ in $S_1$ and you stay there with probability $(1-p)$. If you are in $S_1$, you go back to $S_0$ with probability $(1-p)$ and you go to $S_2$ with probability $p$. If you are in $S_2$, you go back to $S_0$ with probability $(1-p)$ and you go to $S_3$ with probability $p$. If you are in $S_3$, you stay there with probability $1$.
So the entire process is encoded by the transition matrix:
$$ P=\left(\begin{array}{cccc}1-p & p & 0 & 0 \\ 1-p & 0 & p & 0 \\ 1-p & 0 & 0 & p\\ 0 & 0 & 0 & 1\end{array}\right)$$
that is a stochastic matrix, and the wanted probability is $(1\,0\,0\,0)\,P^{12}\,(0\, 0\,0 \, 1)^T$, the probability of ending in $S_3$ after $12$ steps. This particular number can be computed by putting $P$ into its Jordan canonical form: the Jordan base of $P^{12}$ is the same as the Jordan base of $P$ and the eigenvalues of $P^{12}$ are just the twelfth powers of the eigenvalues of $P$.
