Time for arrival is Poisson$(\lambda)$ and serving is Exp$(µ)$. The Markov process is $Z_{t}$ = number of individuals at any time 't'.
When there are no restrictions:
The holding times are $a_{0}$ = $\lambda$ , $a_{1}$ = $\lambda + \mu$ and $a_{i}$ = $\lambda + 2\mu$ when $i\geq2$
and the jump probability will be $P_{i,i+1}$ = $\lambda$/$(\lambda + 2\mu)$ and $P_{i,i-1}$ = $2\mu$/$(\lambda + 2\mu)$ when $i\geq2$
Also, $P_{0,1}$ = 1 and $P_{1,2}$ = $\lambda$/$(\lambda + \mu)$ and $P_{1,0}$ = $\mu$/$(\lambda + \mu)$
This is what I have so far...
How things will change if there is some restriction like there is a limited capacity of let's say 6(there can be almost 6 people in the system including those who are being served/attended) and if someone comes and the Queue is full - they just leave.
When talking about the holding times, I think you mean that they are exponentially distributed with rates $\lambda$, $\lambda+\mu$ and $\lambda+2\mu$, meaning that the mean holding times are $1/\lambda$, $1/(\lambda+\mu)$ and $1/(\lambda+2\mu)$.
If you can have at most 6 people in the system, then the state space for $Z_t$ is $\{0,1,2,\ldots,6\}$. If you are in state 6, you can ignore the arrival process (you will turn away anyone who arrives anyway), and to get out of this state, you need to complete one of the two services. In other words, in state 6, the holding time will be exponential with rate $2\mu$ and the jump probability is $P_{6,5} = 1$.
So in this restricted case, state 6 will be similar to state 0: In state 0, you wait for the next arrival and then jump to state 1. Conversely, when in state 6, you wait for the next service completion and then jump to state 5.