PIC16F57 - problem implementing wake up signal on key stroke
The example circuit is given as this:



The important factor here is the voltage which the \$\overline{MCLR}\$ pin sees, and when.

Under normal operation the voltage should be high enough to register as a logic HIGH. When you want to reset, or wake up, the voltage must be low enough to register as a logic LOW.

With SW1 open, the voltage at point \$A\$, assuming \$V_{CC} = 5V\$, would be (once the capacitor \$C\$ is charged up) 5V.  That is regardless of the value of resistors. Maybe 47KΩ causes too little current to flow to charge the capacitor fast enough? Unlikely, but possible. Try reducing the value of the capacitor (or try temporarily removing it altogether - it's not a component that's critical to the operation of the chip, only the wakeup system).

When you close SW1 with RB2 low the resistors R1, R2 and R7 form a voltage divider. The voltage at point \$A\$ would be \$V_A=\frac{R2 + R7}{R1 + R2 + R7}×V_{CC} = \frac{4800}{51800}×5 = 0.463V\$.

That should be low enough to register as a logic low.  In the datasheet a logic low for \$\overline{MCLR}\$ is defined as \$V_{IL} = 0.15×V_{CC}\$, so for a 5V supply it would be 0.75V.

However, reducing R1 to 4.7KΩ would massively change that voltage:

\$V_A=\frac{R2 + R7}{R1 + R2 + R7}×V_{CC} = \frac{4800}{9500}×5 = 2.526V\$

That's way too high to register as a logic LOW, so a reset will never happen.

So you need to ensure that the ration between R1 and R2+R7 (and of course R1 and R3+R8) is correct.  If you reduce R1 you must also reduce R2 and R3 accordingly.

Maybe going to R1=10KΩ and R2 / R3 = 1KΩ might bring it back into line with the datasheet.