LICQ implies MFCQ and makes KKT multipliers unique

Put the equality-gradient rows and all active inequality-gradient rows into a matrix $M$. By [F1], its rows are independent. If $M$ has no rows, take $w=0$. Otherwise $M{M}^T$ is invertible: for nonzero $\eta$, ${\eta }^{T}M{M}^T\eta =\mathrm{\parallel }{M}^T\eta {\mathrm{\parallel }}_2^2>0$ by row independence. For the target vector $r$ whose equality entries are $0$ and whose active inequality entries are $-1$, set $w=M^T(M{M}^T{)}^{-1}r$. Then $Mw=r$, so $Dh(x)w=0$ and $D{g}_i(x)w=-1<0$ for every active $i$. This is MFCQ.

If $x$ is a local minimizer, [F2] now supplies at least one KKT pair $(\nu ,\lambda )$.

Suppose $(\nu ,\lambda )$ and $(\widehat{\nu },\widehat{\lambda })$ are two KKT pairs at $x$. Complementarity and strict negativity of each inactive $g_i(x)$ force both corresponding multipliers to be $0$. Subtracting the two stationarity equations leaves a linear combination of the equality gradients and active inequality gradients equal to $0$. LICQ makes every coefficient zero, hence $\nu =\widehat{\nu }$ and $\lambda =\widehat{\lambda }$.