Making CO from CO2 is a potential energy source. The value of Kp for the reaction is is 1.50 at 700.0°C. CO2​(g)+C(s)=2CO(g)?

This is a heterogeneous equilibrium. Graphite (C(s)) is solid which does not take place in the equilibrium expression.

Before the reaction occurs :
Reaction quotient = P(CO)² / P(CO₂) = 3.50² / 1.35 = 9.07 > Kp
It favors the backward reaction.

____________ CO₂(g) __+ __ C(s) ⇌ __ 2 CO(aq) __ Kp = 1.5
Initial: _____ 1.35 atm _____________ 3.50 atm
Change: _____ +y atm ______________ -y atm
At eqm: __ (1.35 + y) atm ________ (3.50 - 2y) atm

At equilibrium :
Kp = P(CO)² / P(CO₂)
1.5 = (3.50 - 2y)² / (1.35 + y)
1.5 (1.35 + y) = (3.50 - 2y)²
2.025 + 1.5y = 12.25 - 14y + 4y²
4y² - 15.5y + 10.225 = 0
y = [15.5 ± √(15.5² - 4×4×10.225)] / (2×4)
y = 3.03 (rejected for 3.50 - 2×3.03 < 0) or y = 0.843

Equilibrium partial pressure of CO = (3.50 - 2×0.843) atm = 1.814 atm ≈ 1.81 atm (to 3 sig. fig.)
Equilibrium partial pressure of CO₂ = (1.35 + 0.843) atm = 2.193 atm ≈ 2.19 atm (to 3 sig. fig.)

you need the equation for kp
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Equilibria/Chemical_Equilibria/The_Equilibrium_Constant/Calculating_An_Equilibrium_Concentration_From_An_Equilibrium_Constant/Calculating_An_Equilibrium_Constant%2C_Kp%2C_Using_Partial_Pressures

you need to use an ice table
https://en.wikipedia.org/wiki/RICE_chart
https://www.youtube.com/watch?v=tT-2xk9ZG_A
etc

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you also need to recognize that for ideal gases at constant T and V, the coefficients of the balanced equation can be read as MOLE ratios... AND... as PRESSURE ratios

this
.. 1 CO2(g) + 1 C(s) <----> 2 CO(g)
means
.. 1 mole CO2 yields 2 mole CO
AND
.. 1 atm CO2 yields 2 atm CO

why? from the ideal gas law
.. PV = nRT
.. P/n = RT/V
if T/V is a constant.. (and R is a constant) we can write
.. P1/n1 = RT/V = P2/n2
or
.. P1/n1 = P2/n2
or
.. P1/P2 = n1/n2
i.e..
.. pressure ratios = mole ratios

anyway.. one last hint... ONE possibility is that the reaction is driven to the right..
so that some CO2 is consumed to produce some CO

.. .. .. . . .. .. ..CO2.. .. . .. .. . CO
Initial.. . .. .. ..1.35.. .. .. .. .. ..3.50
Change.. .. ..-X.. .. .. .. . ... .+2X.. . . .. based on coeff of balanced equation, X CO2 ---> 2X CO
Equilibrium..1.35 - X.. ... .. . 3.50 + 2x

so that
.. ... .. . [3.50 + 2x]²
.. Kp = ---- ---- ----- = 1.50
.. .. . .. .[1.50 - X]¹

another possibility is that the reaction is shifted to the left
so that some CO is consumed to produce some CO2

.. .. .. . . .. .. ..CO2.. .. . .. .. . CO
Initial.. . .. .. ..1.35.. .. .. .. .. ..3.50
Change.. .. ..+X.. .. .. .. . ... .-2X.. . . .. based on coeff of balanced equation, X CO2 ---> 2X CO
Equilibrium..1.35 + X.. ... .. . 3.50 - 2x

so that
.. ... .. . [3.50 - 2x]²
.. Kp = ---- ---- ----- = 1.50
.. .. . .. .[1.50 + X]¹

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you get to finish.. solve for X.. calculate values at equilibrium