If the luminosity of a K5V star with 0.70 the mass of the Sun is 0.153 the luminosity of the Sun, it is easy to calculte the distance where the planet will receive exactly as much radiation from the star as Earth gets from the Sun. I call that the Earth Equivalent Distance or EED.
The EED should equal the square root of the star's luminosity relative to the luminosity of the Sun. If the star's luminosity is 0.153 times the luminosity of the Sun, the EED of the star will be at the square root of 0.153, or 0.391152144, times 1 Astronomical Unit, or AU. Thus the EED will be at about 0.391152144 AU.
One AU is defined as 149,597,870,700 meters, which is 149,597,870.7 kilometers, or 92,955,807.273 miles. So if the planet orbits at the EED of the star, it will have the semi-major axis of its orbit at a distance of 58,515,527,860 meters, or 58,515,527.86 kilometers, or 36,359,863.31 miles.
I note that the table in this article:
https://en.wikipedia.org/wiki/K-type_main-sequence_star#:~:text=A%20K%2Dtype%20main%2Dsequence,%2Dtype%20main%2Dsequence%20stars.
says that a K5V star with a mass of 0.70 the mass of the Sun, will have a luminosity 0.17 that of the Sun.
If the star has 0.17 the luminosity of the Sun, the EED will be at 0.412310562 AU. That corresponds to a distance of 61,680,782,230 meters, or 61,680,782.23 kilometers, or 38,326,661.14 miles.
You really should find out which is the more accurate luminosity.
The star has 0.70 times the mass of the Sun. The Sun has a mass of 1.9885 times 10 to the 24th power kilograms, or 1,988,500,000,000,000,000,000,000 kilograms.
I looked for online planetary orbital period calculators.
According to this calculator,
https://www.calculatoratoz.com/en/orbital-period-calculator/Calc-34458
the orbital period at 58,515,527.86 kilometers should be 81.84786 Earth days, and the orbital period at 61,680,782.23 kilometers should be 115.497 days.
According to this calculator,
https://www.calctool.org/astrophysics/orbital-period
the orbital period at a distance of 0.391152144 AU would be 106.75 days, and at a distance of 0.412310562 AU would be 115 days 13 hours, or 115.541666666 days.
The second figure is fairly close to that from the first calculator, but the first one is quite different.
The answer by user177107 to this question:
https://astronomy.stackexchange.com/questions/40746/how-would-the-characteristics-of-a-habitable-planet-change-with-stars-of-differe/40758#40758
Has a table with characterstics of different types of stars. It says a K5V star has a mass of 0.68 Sun, a radius of 0.698 Sun, a surface temperature of 4410 K, and a luminosity of 0.165 Sun. It says that the orbital distance to receive the same amount of radiation as Earth gets from the Sun is 0.406 AU, and the orbital period at that distance is 114.84 days with an orbital speed of 38.518 kilometers per second.
You don't have to put your planet at the exact EED distance from its. The planet can be closer or farther from the star, as long as it is within the star's circumstellar habitable zone.
https://en.wikipedia.org/wiki/Circumstellar_habitable_zone#Solar_System_estimates
Unfortunately, estimates of the circumstellar habitable zone of the Sun to use as a basis for calculating the inner and outer edges of the habitable zonezones of other stars vary greatly. If a science fiction writer is certain they will never need to have another habitable planet in the same solar system they can put their only habitable planet somewhere between 0.99 and 1.01 of the EED of the star to play it safe. But if they want to have several habitable planets in different orbits around the star they need to check various estimates of the Sun's habitable zone to see if a wide habitable zone seems plausible to them.
