Completion and algebraic closure

First you have to observe that since all extensions of the valuation to $\bar{K}$ are conjugate, $\hat{\bar{K}}$ is well-defined up to (non-unique) isomorphism.
Now, since $\hat{\bar{K}}$ is complete and $K$ is dense in $\hat{K}$, the inclusion $K\subset \hat{\bar{K}}$ extends continuously to $K\subset \hat{K}\subset \hat{\bar{K}}$ (in fact you can identify $\hat{K}$ with the closure of $K$ in $\hat{\bar{K}}$). Since $\hat{\bar{K}}$ is an algebraically closed extension of $\hat{K}$, it contains a unique copy of $\bar{\hat{K}}$, namely the algebraic closure of $\hat{K}$ in it. Moreover this copy obviously contains $\bar{K}$ which is dense in $\hat{\bar{K}}$, hence it is also dense. Since $\hat{\bar{K}}$ is complete, it is therefore isomorphic to the completion of $\bar{\hat{K}}$.

The completion of the algebraic closure of a valued field is algebraically closed and complete.

So any further operation of closure or completion gives you a field isomorphic to $\hat{\bar{K}}$.