Let $\ Y_1, Y_2, \dots Y_n \sim Geom(\theta) $ and I set prior for $\ \theta \sim Beta(\alpha, \beta) $
Denote $\ q_k : P\{Y_i \ge k\} $ and I need to find a bayesian estimate with lost MSE for $\ q $ , meaning $\ E[q|y] $
now I know that $\ \theta | y \sim Beta(\alpha +n, \beta + \sum y_i - n) $ and since $\ P(Y_i \ge k ) = (1-\theta)^{k-1} $
so
$$\ E[g(\theta)|y] = E[(1-\theta )^{k-1} | y] = \int_0^1(1-\theta)^{k-1} \pi(\theta|y) d\theta \\ = \int_0^1 (1-\theta)^{k-1} \cdot \theta^{\alpha + n}\cdot (1-\theta)^{\beta +\sum y_i - n} \cdot \frac{1}{B(\alpha + n, \beta +\sum y_i -n)} d\theta\\ = \frac{1}{B(\alpha + n, \beta + \sum y_i -n)}\int_0^1 (1-\theta)^{\beta+\sum y_i -n +k - 1} \cdot\theta^{\alpha+n} d\theta $$
now I guess there is some trick I need to do to turn into integral of density function and then it will just be equal $\ 1 $ but I can't see how ?
In these case it would be better to specify what do you mean with: Geometric distribution. In fact there are two different parametrization for that distribution. Assuming that $P(Y_i\geq k)=(1-\theta)^{k-1}$ as you wrote, I argue that the geometric law you refer to is the one counting the trial before the firs success.
Thus the posterior is, correctly,
$$(\theta|\mathbf{y})\sim \text{Beta}(\alpha+n;\beta+(\Sigma y) -n)$$
at this point you have to calculate
$$E[(1-\theta)^{k-1}|\mathbf{y}]=\int_0^1{(1-\theta)^{k-1}}\pi(\theta|\mathbf{y})d\theta=$$
$$=\frac{\Gamma(\alpha+\Sigma y+\beta)}{\Gamma(\alpha+n)\Gamma(\Sigma y+\beta-n)}\int_0^1\theta^{(\alpha+n)-1}(1-\theta)^{(\Sigma y+\beta-n+k-1)-1}d \theta=$$
$$=\frac{\Gamma(\alpha+\Sigma y+\beta)}{\Gamma(\alpha+n)\Gamma(\Sigma y+\beta-n)}\frac{\Gamma(\alpha+n)\Gamma(\Sigma y+\beta-n+k-1)}{\Gamma(\alpha+\Sigma y+\beta+k-1)}\underbrace{\int_0^1\frac{\Gamma(\alpha+\Sigma y+\beta+k-1)}{\Gamma(\alpha+n)\Gamma(\Sigma y+\beta-n+k-1)}\theta^{(\alpha+n)-1}(1-\theta)^{(\Sigma y+\beta-n+k-1)-1}d \theta}_{=1}=$$
$$=\frac{\Gamma(\sum y-n+\beta+k-1)}{\Gamma(\alpha+\sum y+b+k-1)}\cdot \frac{\Gamma(\alpha+\sum y+\beta)}{\Gamma(\beta+\sum y-n)}$$