The balanced equation is
"HNO"_3 + "KOH" → "KNO"_3 + "H"_2"O"
or
"H"^"+" + "OH"^"-" → "H"_2"O"
"Initial moles of H"^+ = "0.040 00" color(red)(cancel(color(black)("L HNO"_3))) × (0.100 color(red)(cancel(color(black)("mol HNO"_3))))/(1 color(red)(cancel(color(black)("L HNO"_3)))) × ("1 mol H"^+)/(1 color(red)(cancel(color(black)("mol HNO"_3)))) = "0.004 00 mol H"^+
"Moles of OH"^"-" color(white)(l)"added" = "0.025 00" color(red)(cancel(color(black)("L KOH"))) × (0.100 color(red)(cancel(color(black)("mol KOH"))))/(1 color(red)(cancel(color(black)("L KOH")))) × ("1 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol KOH")))) = "0.002 50 mol OH"^"-"
You have more moles of "H"^"+" than of "OH"^"-", so all of the "OH"^"-" will be neutralized.
The moles of "H"^"+" remaining are
"(0.004 00 - 0.002 50) mol" = "0.001 50 mol"
The total volume is now
"(40.00 + 25.00) mL = 65.00 mL = 0.065 00 L"
∴ ["H"^"+"] = "moles"/"litres" = ("0.001 50 mol")/("0.065 00 L") = "0.023 08 mol/L"
"pH" = "-log"["H"^"+"] = "-log(0.023 08)" = 1.64
