First, a non-answer, then a kind-of answer.
BIC - and other information criteria such as AIC - are really meant for model comparison (which involves two different models, not two estimates of the same model) than for model evaluation (which only involves one model.) You can see that by writing out the expression for BIC:
$\text{BIC} = -2\ln L(\hat{\theta};x) + k \ln N$
where $L(\cdot;\cdot)$ is the likelihood function, $k$ is the number of parameters to be estimated, etc. In your case, you are specifying a model with an unknown breakpoint, but you're not comparing it to anything, so far as I can tell. If you attempt to use BIC as a tool for selecting the best estimates, observe that regardless of what the estimates are, the term $k \ln N$ will remain the same, so you will in effect just be maximizing the other term, which is the log likelihood function, and getting the MLE. No need (or value add) for BIC in this case!
Now, if you were comparing two different model formulations, e.g., one with the data split at an unknown to-be-estimated breakpoint $m$ and each part fit to a model with $n_1$ and $n_2$ parameters respectively, and comparing that to a model with no breakpoint but $n_3$ parameters, you'd have $k_1 = n_1 + n_2 + 1$ (for $m$) for the first model and $k_2 = n_3$ for the second model. Judging from the graph above, you have a constraint that forces the fits on the "left" and "right" parts of the data set to intersect at the breakpoint; this constraint would typically remove one free parameter from the parameter count (as the estimates have to satisfy the constraint, and typically this can be done by making one parameter a function of all the others including the location of the breakpoint, so it's not a free parameter any more.) In such a case, $k_1 = n_1 +n_2$, as we subtract 1 from the previous expression for $k_1$ of $n_1 + n_2 + 1$.
However, if you know the breakpoint, you don't add 1 for it to the expression for $k$. A heuristic explanation is that it's not a parameter to be estimated, so it doesn't count against model complexity. There's a real math explanation for this also, but perhaps this isn't the place to go over that.