= Solution
A precise conditional convergence statement is: with the smoothness and <convexity> assumptions of the question, a strict <feasible point>, $\mu_k\downarrow0$, and barrier stationarity residuals
$$
r_k=\nabla f(x^k)-\sum_i\lambda_i^k\nabla c_i(x^k)\longrightarrow0,
$$
every convergent sequence of strictly <feasible points> $x^k\to x_*$ has a <global minimizer> as its <limit>. The <Lagrange multiplier> estimates are bounded and have a <subsequence> converging to multipliers satisfying the <KKT conditions>. The theorem does not assert convergence of the primal sequence without further assumptions.
First, <continuity> gives $c_i(x_*)\geq0$. For an inactive <optimization constraint>, $c_i(x_*)>0$, so
$$
\boxed{\lambda_i^k=\mu_k/c_i(x^k)\longrightarrow0}.
$$
For the <active constraints>, the hint permits assuming $\lambda_i^k\to\lambda_i^*$. Combining this with the inactive limits gives $\lambda_*\geq0$, and passage to the <limit> in the residual equation proves stationarity. Since $\lambda_i^kc_i(x^k)=\mu_k\to0$, <complementary slackness> also holds. All <KKT conditions> follow, and part (b) proves global optimality.
The hint's multiplier assumption can in fact be weakened to selecting a <subsequence>. To see why the estimates are bounded, choose a strict <feasible point> $z$ and put $\delta=\min_i c_i(z)>0$. <Concavity> gives
$$
\delta\|\lambda^k\|_1\leq\sum_i\lambda_i^kc_i(z)\leq p\mu_k+(\nabla f(x^k)-r_k)^T(z-x^k).
$$
The right side is bounded because $x^k\to x_*$ and $r_k\to0$. This is the <bounded barrier multipliers under Slater's condition> argument. A convergent multiplier <subsequence> then gives the same <KKT point>, which suffices to prove that the assumed primal <limit> is globally optimal.
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