## 1 Introduction and Preliminaries
 \tag{11}\]  
and assuming that  
\[m(\mu_{n-1},\mu_{n})\leq\max\{M_{d}(\mu_{n-1},\mu_{n}),M_{d}(\mu_{n},\mu_{n+1} )\}=M_{d}(\mu_{n-1},\mu_{n}),\]  
then inequality (11) can be written as,  
\[M_{d}(\mu_{n},\mu_{n+1})\leq\frac{(1-\epsilon)}{2}\beta(M_{d}(\mu_{n-1},\mu_{n }))M_{d}(\mu_{n-1},\mu_{n})+K_{2}(\epsilon)^{\mu}\psi(\epsilon),\]  
for each \(\epsilon\in[0,1]\). Hence, \(\{\mu_{n}\}\) is a Cauchy sequence in a closed subset \(\mathcal{P}_{0}\) of a complete metric space \((\Omega,M_{d})\). As we know that  
\[M_{d}(\mu_{n+1},T\mu_{n})=M_{d}(\mathcal{P},\mathcal{Q}),\]  
and every Cauchy sequence in a complete metric space is convergent, there exists some \(\mu\in\mathcal{P}_{0}\) such that \(\mu_{n}\to\mu\) in \(\mathcal{P}_{0}\subseteq\mathcal{P}\),  
\[\lim_{n\to+\infty}M_{d}(\mu_{n},\mu)=0.\]  
Hence, \(\mathcal{P}_{0}\) is \(\alpha\)-\(\theta\)-proximal orbital complete. **Theorem 1**. : _Let \(\mathcal{P}\) and \(\mathcal{Q}\) be nonempty subsets of a complete metric space \((\Omega,M_{d})\). Furthermore, assume that the subset \(\mathcal{P}\) is closed and we have a continuous mapping \(T:\mathcal{P}\to\mathcal{Q}\), which is an \(\alpha\)-\(\theta\)-proximal admissible and \(\alpha\)-\(\theta\)-Geraghty-Pata-Suzuki-type proximal contraction. Additionally, let \(T(\mathcal{P}_{0})\subseteq\mathcal{Q}_{0}\), where \(\mathcal{P}_{0}\
eq\phi\). If \(\mathcal{Q}\) is approximately compact with respect to \(\mathcal{P}\), then \(\mu\) is a unique best proximity point for \(T\) in \(\mathcal{P}_{0}\)._  
Proof. : Consider \(\mu_{0}\) in \(\mathcal{P}_{0}\), from Lemma (4), the sequence \(\{\mu_{n}\}\) is a Cauchy \(\alpha\)-\(\theta\)-proximal admissible Picard sequence in \(\mathcal{P}_{0}\). As a result, we can derive a sequence \(\{\mu_{n}\}\) in \(\mathcal{P}_{0}\) such that  
\[\left. \begin{array}{l}\alpha(\mu_{n},\mu_{n+1})\geq\theta(\mu_{n},\mu_{n+1}) \\ M_{d}(\mu_{n},T\mu_{n-1})=M_{d}(\mathcal{P},\mathcal{Q})\\ M_{d}(\mu_{n+1},T\mu_{n})=M_{d}(\mathcal{P},\mathcal{Q})\end{array} \right\} \text{implies }\alpha(\mu_{n-1},\mu_{n})\geq\theta(\mu_{n-1},\mu_{n}),\]  
for all \(n\in\mathbb{N}\). Given that \(T\) is an \(\alpha\)-\(\theta\)-proximal admissible mapping, and it is also an \(\alpha\)-\(\theta\)-Geraghty-Pata-Suzuki-type proximal contraction, we can apply reasoning similar to that used in the proof of Lemma (4). Consequently, we can deduce that the sequence \(\{\mu_{n}\}\in\mathcal{P}_{0}\) is a Cauchy \(\alpha\)-\(\theta\)-proximal admissible Picard sequence. Since \(\Omega\) is complete and \(\mathcal{P}\) is closed, there exists some \(\mu\) in \(\mathcal{P}\) such that \(\lim_{n\to+\infty}M_{d}(\mu_{n},\mu)=0\). Since every Cauchy sequence in a complete metric space converges, there exists a \(\mu\) in \(\mathcal{P}_{0}\) such that \(\mu_{n}\to\mu\) within \(\mathcal{P}_{0}\). Furthermore, the mapping \(T\) is continuous, so we have \(T\mu_{n}\to T\mu\). Now, based on the above inequality, we have:  
\[M_{d}(\mathcal{P},\mathcal{Q})=\lim_{n\to+\infty}M_{d}(\mu_{n},T\mu_{n})=M_{d}( \mu,T\mu).\]  
Thus, we can deduce that \(\mu\) represents the \(\alpha\)-\(\theta\)-Geraghty-Pata-Suzuki-type best proximity point of the mapping \(T\).