how to convert if-else with multiple condition with design pattern C#

i am aware of how to to convert if-else with single conditions in each if with design pattern. but now i am little bit confuse because in my code there are multiple condition separated by && operator in if. now how to refactor such type of cases.

code is as per below:

    public ActionResult PaymentDetails(PaymentlistingDTOVM model, string SearchButton)
    {
        PaymentlistingDTOVM dto = new PaymentlistingDTOVM();

        if (model.PaymentSearch == null)
        {
            ModelState.AddModelError("Message", "Please select type to search ( Member No / Member Name).");
        }
        else if (model.PaymentSearch == "1" && model.Memberno == null)
        {
            ModelState.AddModelError("Message", "Please Enter Member No.");
        }
        else if (model.PaymentSearch == "2" && model.MemberName == null)
        {
            ModelState.AddModelError("Message", "Please Enter Member Name.");
        }
        else
        {
            if (!string.IsNullOrEmpty(model.SearchButton) || model.Page.HasValue)
            {
                if (model.Memberno != null)
                {
                    var Listpay = objIPaymentlisting.AllPaymentDetails(Convert.ToString(model.Memberno));

                    var results = Listpay.Where(p => (p.MemberID == model.Memberno));

                    var pageIndex = model.Page ?? 1;

                    model.SearchResults = results.ToPagedList(pageIndex, RecordsPerPage);
                }
                else if (model.MemberName != null)
                {
                    string[] Memno = model.MemberName.Split('|');

                    var Listpay = objIPaymentlisting.AllPaymentDetails(Convert.ToString(Memno[3]));

                    var results = Listpay.Where(p => (p.MemberID == Convert.ToInt32(Memno[3])));

                    var pageIndex = model.Page ?? 1;

                    model.SearchResults = results.ToPagedList(pageIndex, RecordsPerPage);
                }
            }
        }

        return View(model);
    }